in this video we're going to focus on problems associated with gibbs free energy so let's start with this question which of the following statements is false the change in free energy delta g is less than zero for a spontaneous process is that true or false and what about b is it equal to zero at equilibrium or is d true is it greater than zero for a spontaneous process so what you need to know is that when delta g is basically positive or you could say when it's greater than zero this is true for a non-spontaneous process delta g is equal to zero during equilibrium when you have a reversible process and when delta g is negative or when it's less than zero this process is spontaneous now a reaction proceeds spontaneously when it could lower its energy in fact a natural spontaneous process will occur in such a way to find the lowest possible energy state and so we can see that when a change in free energy is negative it's usually it's always associated with a spontaneous process and so that's something you want to keep in mind so which answers are true and which ones are false so let's look at a the change in free energy is less than zero for a spontaneous process and that's true looking at b the change in free energy is equal to zero at equilibrium that's also true so we're looking for the false statement and c the maximum amount of work that can be obtained from a spontaneous process is equal to the change in free energy that is also true the maximum work that can be obtained is equal to delta g for a spontaneous process now if you have a non-spontaneous process to make it work you got to put energy into it and so the minimum work required to drive a non-spontaneous reaction forward is also equal to delta g now we use the word minimum because you need to put in energy to overcome the energy losses that occur due to friction so if you're trying to drive a non-spontaneous process forward you need more energy than delta g so the minimum amount to get it started is equal to the change in free energy but if you have a spontaneous reaction where you're trying to get energy from the reaction then the maximum energy that you can get or the maximum useful work that it can do is equal to the change in the free energy so c is the true statement now d the change in free energy is greater than zero for a spontaneous process it's greater than zero for a non-spontaneous process so d is the false statement which means d is the answer that we're looking for number two the change in enthalpy and entropy for a certain reaction are those values calculate the change in free energy at 25 degrees celsius so how can we do this what formula do we need the change in free energy also known as gibbs free energy is equal to the enthalpy change minus the temperature multiplied by the entropy change now you need to be careful with the units so notice that delta h the enthalpy change is in kilojoules per mole and delta g is also in kilojoules per mole but notice that the entropy change is in joules per mole per kelvin so we need to convert joules to kilojoules and to do that all we need to do is divide by a thousand so let me show you the conversion so we have 212 joules per mole per kelvin and we know that there's a thousand joules per one kilojoule so we just got to divide this number by a thousand so the entropy change is going to be 0.212 kilojoules per mole per kelvin so just keep that in mind so now let's work on this problem so delta h is negative 46.5 kilojoules per mole now the temperature has to mean kelvin so 25 degrees celsius plus 273 that's 298 kelvin and the entropy is 0.212 kilojoules per mole per kelvin so notice how the unit kelvin cancels and so now this quantity is in kilojoules per mole and this quantity also is in kilojoules per mole so now let's get the answer so it's negative 46.5 minus 298 multiplied by positive 0.212 and so delta g in this example is equal to negative 109.7 kilojoules per mole now is the reaction spontaneous non-spontaneous or equilibrium what would you say whenever delta g is negative or when delta g is less than zero what we have is a spontaneous process and so that's the answer to the second part now this value corresponds to answer choice d which is rounded to negative 110 and so that's it for this problem number three calculate the free energy change for the reaction shown below at 25 degrees celsius so what we need to do is we need to calculate delta g for that reaction and we could use this formula it's going to be delta h minus t times delta s now using the information in this table we could calculate delta h for the reaction and delta s and once we have that we can plug it into this formula to get delta g so let's go ahead and do that let's calculate the enthalpy change for the reaction these values that you see here they represent the enthalpy change of formation for each of those substances and those values will always be zero for a pure element in its standard state now the formula that we're going to use is this formula we're going to take the sum of the moles times the heat of formation for each reactant and product but first this is going to be for the products and then minus the sum of the moles times the heat of formation of all the reactants so that's the formula we need to use so let's start with the products all we have is a 2 in front of hcl so it's going to be 2 moles of hcl times the heat of formation for hcl which is negative 92 kilojoules per mole and then for the reactants so for h2 we have one mole of that the enthalpy change for h2 is zero for cl2 it's one mole that's the coefficient and then times zero as well two times negative 92 that's going to be negative 184 and the unit moles will cancel so what we have here is the delta h for the reaction is going to be negative 184 kilojoules so we'll save that answer and now let's calculate the entropy change for the reaction so we want to get delta s at this point and we're going to use the similar process so we're going to take the sum of the product of the moles times the entropy values for each product and then we're going to take the sum of the moles of the reactants times the entropy of each reactant so we're going to have 2 moles of hcl and the entropy for hdl is 187. the units be in joules per mole per kelvin so once again the unit moles will cancel which will give us the entropy change in joules per kelvin now let's focus on the reactants so we have one mole of h2 and the entropy value for that is 131 and then one mole of cl2 times the entropy value for that which is 223 joules per mole per kelvin so in all cases the unit moles will cancel so this is going to be 2 times 187 minus 131 minus 223 thus the entropy change for the reaction is going to be positive 20 joules per kelvin now let's go ahead and calculate delta g since we now have everything that we need delta g is going to be equal to delta h minus t times the delta s so delta h in this example is negative 184 kilojoules the temperature is 25 degrees celsius and if we add 273 to that that's going to be 298 kelvin and then we're going to multiply that by the entropy change of not joules but kilojoules so we need to convert 20 joules into kilojoules if we divide that by a thousand that's going to be positive 0.02 kilojoules per kelvin so the unit kelvin will cancel and we'll get delta g for the reaction in kilojoules and this is going to be standard delta g because we use standard delta h and standard delta s so it's negative 184 minus 298 times 0.02 so this is going to be negative 189.96 kilojoules so we could say that is approximately negative 190 kilojoules therefore answer choice a is the correct answer in this example number four calculate the free energy change of the reaction shown below using the standard free energy change of formation values in the data table so this is going to be similar to the last problem but we can calculate delta g directly using the table rather than the formula like we did in the last problem so the free energy change for the reaction is going to be the sum of the moles times the delta g values these are the standard detail the standard delta g values of formation for all the products and then minus the sum of the moles of the reactants times the standard delta g values of formation for the reactants so let's start with the products so we have two moles of iron oxide and the value for that is negative 740 kilojoules per mole now for the reactants we have four moles of iron two oxide and we're going to multiply that by negative 255 kilojoules per mole now oxygen is a pure element in its standard state so therefore the delta g of formation for that is going to be zero so we could just put plus one mole times zero so once again the unit moles will cancel and we're going to get delta g in kilojoules so we have two times negative 740 minus negative 255 times four so i got negative 460 kilojoules so be careful with this double negative here there's a negative sign and another one there so make sure you incorporate that that's the correct answer it's going to be answer choice d now i want to take a minute to mention something when dealing with thermochemical equations the energy change associated with a reaction that is the amount of heat energy absorbed or released it's typically in units of kilojoules so for this first reaction let's say it releases negative 20 kilojoules of thermal energy if we multiply the coefficients by two and then we're gonna get double the amount of energy released because we we've doubled the quantity of reactants that are converted into products if we multiply the original equation by three the energy change will increase by a factor of three now sometimes instead of seeing kilojoules as the energy associated with the energy change that is that occurs for a reaction you might see kilojoules per mole and when you see that that's basically the amount of energy that's absorbed or released per mole of reaction if we were to take negative 20 divided by one or one mole of a or you could think of it as just one mole of the reaction the energy change will be negative 20 kilojoules per mole now this ratio won't change regardless of what the coefficients of the reaction will be so if you take negative 40 divided by 2 you're going to get negative 20 still if you take negative 60 divided by 3 you're going to get negative 20. and when calculating the non-standard delta g value it's important that we use this figure as opposed to these other numbers and here's why late in this video we're going to use this equation the non-standard delta g value is equal to the standard delta g value plus rt lnq where q is the reaction quotient r has the units joules per mole per kelvin delta g we can have that in kilojoules per mole or joules per mole temperature is in kelvin so in order for this equation to work in order for us to algebraically add these quantities they need to be in the same units so if we were to report delta g naught as being in kilojoules and not kilojs per mole it won't work well with this equation so we need it to be in joules per mole and not just joules we can convert kilojoules from moles to juice from all but if we just have kilojoules and convert it to joules we can't add these two quantities because we need them to be in joules per mole so going forward instead of just leaving the answer in kilojoules let's say from calculating delta h or delta g of the reaction i'm going to be using kilojoules per mole so that it works well with this equation so i want you to be aware of those units and uh their relationship to each other but for this equation to work just make sure that delta g naught is in joules per mole and not just joules or kilojoules number five calculate the free energy change of the decomposition of dinitrogen pentoxide into dinitrogen tetroxide which is n2o4 and oxygen so how can we use the first two equations and free energy associated with those equations to get the free energy of the third equation what do we need to do well we need to change these two equations in such a way that when we add them we get this equation and so we can add the adjusted free energy change values to get this one so how do we need to change equation one and two so that it can be added to this equation well the first thing we need to realize is that notice that in a second equation n2 o4 is in the right spot and we have the right number of it so we don't need to change the second equation now the first equation we do need to change notice that n2o5 is found on the right side but we need it on the left and we need the coefficient to be one so therefore the first reaction we need to reverse it and we need to multiply by half so half of four and o2 is going to be two no2 and then we'll first need to reverse it so let me do that first so half of this will be 1 so that's 1 and 205 and then on the other side half of 4 no2 that's going to be 2 no2 and then half of o2 we can simply write that as 1 half o2 now because we reversed it it's going to change from positive sixty to negative sixty and we multiplied it by half so half of negative sixty is negative thirty so that's the new free energy change for this reaction now this one we're not going to change so i'm just going to rewrite it the way it is and so that's going to stay negative 6. so now let's add these two reactions so notice that nl2 will cancel it's the intermediate in this reaction and so we're left with just 1 and 205 molecule it turns into 1 and 204 molecule and half of an o2 molecule so now all we need to do is add up these two values negative 30 plus negative 6 is negative 36 kilojoules per mole so therefore this is the final answer so answer choice c is the correct answer in this example number six the standard free energy change for a reaction under certain conditions is negative 65 kilojoules per mole at this instant which of the following events will occur so will the free energy change increase or decrease and will the reaction shift to the left or to the right or is the system at equilibrium so let's say if we have two species a and b now if delta g is negative what's going to happen is the reaction spontaneous or non-spontaneous anytime delta g is negative or less than zero the reaction is spontaneous so if we have a spontaneous process what does that mean what's happening in terms of the reaction is the reaction going towards the right or is it going towards the left when delta g is negative you need to know that it's spontaneous in the forward direction so that means that when delta g is negative the reaction is going towards the right it's shifting to the right so it's not going to the left so a is out and c is out and it's not at equilibrium delta g will be at equilibrium when delta g is zero i mean the reaction will be at equilibrium one delta g zero that's what i'm going to say so we're between b and d now as the reaction shifts towards the right will the free energy change increase or decrease right now at this instant the free energy change is negative 65 kilojoules per mole as the reaction shifts towards the right the products will increase and the reactants will decrease and this will continue to happen until it reaches equilibrium and at equilibrium the free energy change will be zero so the reaction will proceed in such a way until delta g becomes zero until the system reaches equilibrium so to go from negative 65 to zero will the free energy change will it increase or decrease well negative 65 is a lower number on the number line relative to zero so let's say if you draw a number line here's zero here's 50 here's negative 65. the lower number is on the left side so therefore delta g has to increase from negative 65 to zero as the reaction proceeds to the right going towards equilibrium so therefore we can eliminate answer choice d it's not going to decrease it's going to increase number seven estimate the boiling point of bromine now we're given the enthalpy of vaporization of bromine it's 30.9 kilojoules per mole and the standard entropy values for liquid and gaseous bromine are 152 and 245 joules per mole per kelvin now let's write a reaction so we have liquid bromine turning into a gaseous bromine and so the enthalpy of vaporization is basically the enthalpy of this reaction because liquid bromine is being vaporized into gaseous bromine so the enthalpy of the reaction is positive 30.9 kilojoules per mole now how can we calculate the boiling point of bromine what do we need well we need to start with this equation delta g is equal to delta h minus t delta s and at the boiling point these two physical states coexist so what that means is that this system is an equilibrium at the boiling point the liquid state and the gaseous state they coexist they can go back and forth at that temperature now if the system is in equilibrium what is delta g at equilibrium delta g is equal to zero so now what we have is zero is equal to delta h minus t delta s so let's take this term and move it to the other side so t delta s is equal to delta h and now we need to calculate t and t represents the boiling point temperature and so to calculate it it's simply the enthalpy of vaporization divided by the entropy change so that's how you can calculate the boiling point of a liquid now we already have the enthalpy change what we're missing is the entropy change so to calculate the entropy change for the reaction it's going to be the sum of the entropy values for the products minus the sum of the entropy values for the reactants now the only product that we have in this reaction is gaseous bromine so i'm going to write g for gas and a reactant is liquid bromine and the entropy value for gaseous bromine is 245 and for liquid bromine it's 152. so let's subtract those values 245 minus 152 and so the entropy change is 93 and the units are joules per mole per kelvin so now we have the entropy change and we have the enthalpy change so with that information we can go ahead and calculate the temperature now we need to be careful the units as you can see the enthalpy change is in kilojoules per mole the entropy change is in joules per mole so i'm going to convert this to kilojoules by dividing it by a thousand so the temperature is going to be the enthalpy of vaporization which is 30.9 kilojoules per mole divided by the entropy change 93 divided by 1000 that's 0.093 and it's going to be kilojoules per mole per kelvin so we can see that the unit kilojoules will cancel and the unit moles will cancel and so we're going to get the temperature in kelvin so it's 30.9 divided by .093 now this is equal to 332 point 26 kelvin and if we subtract this by 273.15 kelvin we can convert it to celsius so the temperature is about 59.1 degrees celsius so we can route it and say the boiling point of bromine is approximately 59 degrees celsius therefore c is the right answer in this problem number eight which of the following statements is false is it a b c d or e well let's go over some things that you need to know now when delta g is less than zero that means it's negative we have a spontaneous reaction and so that means that the reaction shifts to the right if the standard delta g value is equal to zero then that means that the system is at equilibrium and if the standard free energy change is greater than zero then we have a non-spontaneous reaction it's non-spontaneous in the forward direction which means it's spontaneous in a reverse direction so it shifts to the left now keep in mind when delta g is less than zero it's equivalent to being negative and when delta g is greater than zero it's equivalent to being positive now when we have a spontaneous reaction when delta g is less than zero then k is greater than one and at equilibrium k is equal to one and if we have a non-spontaneous reaction k is less than one but it has to be greater than zero k cannot be a negative number now when k is greater than one it's product favored keep in mind k is the ratio of the products to the reactants so let's say if k is a significantly large number like 10 000 that means that at equilibrium there's a lot more products than reactants so any time the reaction shifts to the right for a spontaneous process it means that it's product favored if delta g is positive where k is small like 1 times 10 to the negative 8 or something you have a non-spontaneous process so it's spontaneous in the reverse direction so it's reactant favored when k is small but it's product favored when k is large so if we look at e the reaction is reactant favored when k is significantly less than one that's true and d it's product favored when k is significantly larger than one that's also true answer choice c if the standard free energy change is zero then k is equal to one that is a true statement and for b for a non-spontaneous process k is between zero and one and that's just something you need to know so make sure you write that down in your notes now a has to be the false statement for a spontaneous process k is not less than one in fact k is greater than one when spontaneous because delta g is negative is less than zero so a is the correct answer it's the statement that is false number nine the equilibrium partial pressure constant is one point four times ten to negative five at 298 kelvin for a certain reaction calculate the standard free energy change at 298 kelvin for that reaction so we're given kp the equilibrium partial pressure constant and it's equal to this value so how can we use this number to calculate delta g now there is a formula the standard free energy change is equal to negative rt times the natural log of the equilibrium constant so r in this example is the energy constant 8.3145 joules per mole per kelvin and the temperature in kelvin is 298 kelvin and so we can see that the units kelvin will cancel and k is unitless so we don't have to worry about that now the units that we're going to have is joules per mole and delta g has to be in kilojoules per mole so we need to convert it to that so let's take negative 8.3145 multiply by 298 times the natural log of 1.4 times 10 to the negative 5. so the standard free energy change for this reaction is positive 27 700 joules per mole if you round it now let's convert it to kilojoules per mole so one kilojoule is one thousand joules and now we can cancel this unit so 27 700 divided by a thousand is 27.7 kilojoules per mole and so that's delta g in this problem which means that b is the right answer now let's think about this notice that k is very small it's point zero zero zero zero one four so k is greater than zero but it's less than one which means that we have a reactant favored reaction and notice that delta g is positive we said that if k is less than one delta g is positive or delta g is greater than zero which means that it's non-spontaneous in the forward direction but it is spontaneous in a reverse direction so it's reactant favored it goes towards the left number 10 the standard free energy change for a certain reaction at 298 kelvin is negative 215 kilojoules per mole calculate the value of the equilibrium constant for this reaction so how can we do this what should we do now notice that delta g is equal to a negative number and we know that when it's negative the reaction is spontaneous it's going to shift to the right and it's a product favored reaction now when delta g is negative is k very large or very small if it's product favored k is going to be very large so k is going to be a positive number and it's going to be 10 raised to some positive exponents a large number on the other hand when delta g is positive we have a non-spontaneous process and so the reaction is reactant favored and so k is going to be less than 1 but greater than 0. so because it's greater than 0 is still a positive number but it's less than 1 so the exponent has to be negative as opposed to being positive but for this problem delta g is clearly a negative value so we need a positive k value with a positive exponent we can eliminate answer choice a because it has a negative exponent which means that delta g is positive this also has a negative exponent now k can never be negative so b is out so it's either between c or d that's when k is positive and it has a positive exponent but let's get the exact answer now let's start with this equation delta g is equal to negative rt ln k so the first thing we need to do to isolate k is to divide both sides by negative rt and so ln k is equal to negative delta g over rt now we need to get k by itself when dealing with logarithms here's some general things you need to keep in mind a raised to the c is equal to b so that's how you can convert a logarithmic expression into its exponential form now the base of a natural log is e so e raised to everything on the right side is equal to k or we could say k is equal to e raised to the negative delta g over rt so that's the formula that we're going to use to calculate k so i'm just going to rewrite it here now when using this formula because r is in joules per mole delta g has to be in joules per mole so you got to convert kilojoules to joules so let's do that first so we have negative 215 kilojoules per mole and we know that one kilojoule is equal to a thousand joules so this time we need to multiply by a thousand as opposed to divide by a thousand so it's negative 215 000 joules per mole so k is going to be e negative 215 000. now that's delta g and keep in mind there's another negative sign so it's negative delta g so we've got two negative signs in this example and then r which is 8.3145 multiplied by the kelvin temperature 298. don't plug in a celsius temperature you won't get the right answer and so this is equal to 4.8 times 10 to the 37 so that's the value of the equilibrium constant so c is the answer for this problem number 11 which of the following statements is true concerning the physical reaction shown below is the reaction spontaneous at high temperatures low temperatures is it always spontaneous never spontaneous or is it spontaneous at the freezing point so in this reaction we have liquid water turning into ice so which statement is true if we take a step back and keep it simple the best way to turn liquid water into ice is to put it in the freezer if you put it in the freezer heat energy is going to flow away from the liquid water and so as water loses thermal energy the temperature decreases and as the temperature decreases it's going to freeze to ice so therefore we know that it's spontaneous at low temperatures so just by intuition you know that the answer has to be b but now let's talk about why that's the case so first this reaction is it endothermic or exothermic the freezing of water in order to freeze water in order to go from a liquid to solid you need to remove energy so liquid water has to release heat energy into the surroundings so let's say we have a beaker filled with liquid water if we put it in a freezer with let's say the temperature is negative 20 degrees celsius and the temperature of this water is 25. heat flows from hot to cold so heat is gonna leave the system and it's going to flow into the surroundings whenever the system releases heat energy it's an exothermic process so delta h is negative now the next thing we need to analyze is the entropy change is it positive or is it negative so which one has more entropy a solid or liquid a solid is very organized so it doesn't have much disorder so a solid has a very low entropy value a liquid has a relatively higher entropy value than the solid so we're going from a high entropy value to a low entropy value so entropy is decreasing therefore delta s is also negative now you need to be familiar with this table delta h delta s t and delta g you may want to commit this to memory so when delta h and delta s are both positive or both negative the sine of delta g is temperature dependent so when they're both positive and the temperature is high delta g is going to be negative which means that it's spontaneous under these conditions it's spontaneous at high temperatures and it's non-spontaneous at low temperatures now when delta h and delta s are both negative it's still going to be temperature dependent this time delta g is negative at low temperatures so it's going to be spontaneous at a low temperature but not spontaneous at a high temperature now when the signs of delta h and delta s are different then it's no longer temperature dependent so when delta h is negative and delta s is positive delta g is always spontaneous regardless of the temperature and when delta h is positive and delta s is negative it's always going to be non-spontaneous so in this example delta h and delta s are both negative so we're dealing with this section here so the reaction is going to be spontaneous only at low temperatures and it makes sense liquid water will only freeze at low temperatures so that's why b is the answer we can eliminate a it's not going to be spontaneous at high temperatures and c and d doesn't apply now what about e at the freezing point that is if the temperature is 0 degrees celsius then it's going to be at equilibrium liquid water and ice can coexist at zero degrees celsius so at this temperature delta g will equal zero and so it's going to be at equilibrium now if we decrease the temperature below zero then it's going to shift in this direction it's going to freeze so at low temperatures that is below the freezing point it becomes spontaneous now if we switch if we increase the temperature above zero then it becomes not spontaneous which means that it's spontaneous in a reverse direction so at high temperatures water or ice rather is going to melt into liquid water at high temperatures but at low temperatures liquid water is going to freeze into ice and so you can figure this out based on this table or just using simple intuition you