Hello Assalamualaikum Hello little brothers and sisters meet again with Brother on the kinematics channel in this video we will learn to determine the change in enthalpy of reaction or delta H based on calorimeter data a calorimeter is a tool in which there is a system and there is an environment the system is a chemical reaction or a place where a chemical reaction occurs while the environment is what is around the reaction meaning water and the walls of the calorimeter Okay this is an example of a calorimeter chart yes usually The reaction that occurs in the calorimeter is an exothermic reaction or a reaction that releases heat So if an exothermic reaction occurs in the system then the heat released will be absorbed by the environment namely Water and the walls of the calorimeter well this is in accordance with the black principle that the heat released is the heat absorbed or received by the environment that releases Heat is a reaction makaci release we replace with kirek si while the one that receives Heat is the environment namely Water and the walls of the calorimeter then we accept we replace with Kyai or solution plus Ki calorimeter Why here the negative sign to show that the reaction releases heat or vice versa then Kiai the formula is MC Delta t and Q calorimeter the formula is C deltate where Q is the amount of heat the unit is Joule or call m is the mass of water or solution, the unit is gram then the small c means this one yes it is the specific heat of water or solution, the unit is joule per gram Kelvin or joule per gram degrees Celsius then the large one is the heat capacity of the calorimeter, the unit is jual.per Kelvin or joule per degree Celsius then the last one is deltate deltate is the change in temperature or teacher minus the initial t then How to determine the enthalpy change of reaction using this calorimeter data well if what is known is the specific heat or small C we can immediately use the formula delta h = min MJ deltate pernoll Well here we assume the heat capacity of the calorimeter or C the amount is ignored but if the heat capacity of the calorimeter is also known then to find Delta we can only use the formula delta h = negative MJ deltate plus c.at per mole C for more Okay let's go straight to the example question yes The first question at a temperature of 25° Celsius reacted 50 ml of 0.4 molar HCL with 50 ml of 0.1 molar NaOH in a plastic calorimeter it turns out the temperature rose to 28° Celsius if the density of the solution is 1 gram per ML and the specific heat of the solution is 4.2 Joules per gram Kelvin then the enthalpy of neutralization is to answer this question we immediately determine the formula that we will use note Is the specific heat of the solution known in this question and the heat capacity of the calorimeter is known we see yes Well here what is known is only the specific heat of the solution which is 4.2 Joules per gram Kelvin while the heat capacity of the calorimeter is unknown meaning we use the formula Elta khas = Min McD ltt formal after we determine the formula we immediately enter the data yes the first is M or the mass of the solution in this question the mass of the solution is unknown but what is known is the volume which is 50 ml HCL and 50 ml NaOH then the total volume of the solution is 50 plus 50/100 ml then because what is requested is mass means the volume of this solution we first change it into mass by using its density here it is known that the density of the solution is one gram per ML so the problem of Rutan is the same as the density or ro times the volume is that one times 100 open to 100 grams Hi Keh we enter the data so delta H = negative 100 times c c is the specific heat the specific heat is also known in the question, namely 4.2 Joules per gram Calvin means multiplied by 4.2 then multiplied by Delta teh dalte is the change in temperature well the temperature at the beginning was 25° Celsius then rose to 2018 fell Celsius then the change in temperature or Delta question is equal to 28 minus 25 you know = 3 then we replace Delta question with three yes then per mole well Mal Who do we use because this is the enthalpy of neutralization or neutralization reaction then the mal we use is Mal water So if HCL acid is reacted with NaOH it will produce water and salt we make the reaction Then to find out Mal from H2O we have to find first Mal from acid and Mal from wet earlier in the question it is known that HCL is 50 ml and its concentration is 0.4 molar means the number of moles we just multiply the volume times the molarity which is 50 times 0.4 or = 20 millimol for NaOH is the same yes the volume is the same concentration means enoha is also 2000 remember the mole ratio is the same as the coefficient ratio here the profession is one each yes means H2O also has a number of moles of 20 minimum so mal here we replace it with 20 Well because this is still minimal means we have to change it first to Mal Yes we just multiply it by 10 to the power of min 3 or we divide it by 1000 Okay continue counting let's scribble first yes Zero above one we cross out below one we cross out then we remove the comment means here the zero is finished 42 divided by two that = 21 21 times three 6310 to the power of negative 3 we move it up So 10 ^ 3 the unit is joules per mole we can change the unit to kilojoules means we divide by 1004 the final result is Min 64 Mal It's quite clear yes Now we move on to the second example question as much as 33 ml of 1.4 m hno3 solution is mixed with 42 ml of KOH solution 1m in a plastic calorimeter the initial temperature of the HNO3 and KOH solutions is 25° Celsius each and after being opened the temperature of the solution rises to 30° Celsius if the heat capacity of the calorimeter is ignored while the specific heat of the solution is considered the same as the specific heat of water which is 4.2 Joules G min 1 comment one then the enthalpy of neutralization is Okay, the same as the previous question, let's see first whether the heat capacity of the calorimeter and its specific heat are known Well, this question We have seen that the heat capacity of the calorimeter is ignored and what is known is the specific heat of water which is 4.2 Joules per gram Kelvin Well, that means we use the formula delta h = min McD ltpr mall-mall, we just enter the data m is the mass of the solution in the question the mass is unknown but the volume is known, namely a total of 33 ML plus 42 ML or equal to 75 ml because what we need is the mass, it means we just multiply the volume by the density of the density we assume one, it means the mass of the solution is equal to 75 times one or equal to 75 grams directly we put it into the formula so delta H = minus 75 times c or specific heat see the specific heat in the question is 4.25 lu multiplied by deltate Delta ask to see in the question initially the temperature was 25° Celsius then it rose to 30° Celsius then Delta t = 32 minus 25 or = 7 so we you guys 7 again okay continue the target look for mal-mal who we are looking for because this is a neutralization reaction means Mal water Well water is obtained from the reaction between hno3 and Koh so we make the reaction first then we determine the moles of each of the acid and base for hno3 means 33 times 1.4 molar which is 46.2 millimoles continue molka Oh means 42 times 1 M or = 42 millimoles then How many moles of h2oh because the coefficients are each one means Mal of H2O follows the limiting malpractice or the reactant that has completely reacted, namely 42 millimoles we immediately enter it into the formula so here per 42 times 10 to the power of min 3 Mal okay Let's just calculate 4.25 and 42, we can cross it out above, it becomes 0.1, yes, 75 times 0.1 times 7 = 52.5, now we raise 10 to the power of 3 to 10 ^ 3, the unit is joules per mole, we change it to kilojoules, which means dividing it by 1000. So the result or enthalpy of neutralization of the reaction is Jin 52.5 kilojoules per mole. Okay, edited to make it clearer, let's try one more question. A bomb calorimeter contains 250 ml of water at a temperature of 25° Celsius, then 200 mg of methane gas is burned. The highest temperature reached by water in the calorimeter is 35° Celsius if the heat capacity of the calorimeter is 75 joules per degree Celsius and the specific heat of water is 4.2 Joules per gram of degree Celsius. What is the enthalpy of combustion of methane gas? Now, pay attention to the question, the heat capacity of the calorimeter is known to be 75 joules per degree Celsius and the specific heat of the water is also known. which is 4.2 Joules per gram degrees Celsius because both are known then to find Delta we only use the second formula that is delta H = min McD altatek plus C deltate Permai let's fill in the data directly yes The first is M or the mass of the solution or the mass of water see in the question the mass of water is unknown what is known is the volume of water which is 250 ml Because the density of water is 1 means the mass of water is equal to the volume of water which is 250 grams then multiplied by the specific heat which is 4.2 then multiplied by deltate we see the temperature change from 25° Celsius to 30° Celsius means Delta t is 35 minus 25 you know = 10 but here it is multiplied by 10 then added the heat capacity of the calorimeter which is 75 times deltate Delta question is 10 well then divided by Mal because this is a combustion reaction means the mal we use is the mole of the substance being burned here what is burned is methane gas so we find the mal of methane gas using the formula grams per Mr methane earlier the question mentioned as much as 200 mg or zero point Mba G divided by Mr Mr ch4 = 16 of C12 h10 my4 yes 0.2 divided by 16 = 0.0125 Mal Keh we immediately enter it into the delta h formula 0.0125 we just write 125 times 10 to the power of min 4 Heh we immediately calculate 250 times 4.2 times 10 = 10-5 hundred plus 75 times 10 or 750 divided by 125 times 10 to the power of min 4 then 10500 plus 750 = 11250 10 to the power of min 4 we raise it up to 10 ^ 4/125 then the final result is min90 times 10 to the power of 4 joules per mole Hi or = Min 900 kilo joules per mole Okay that's all for this video, thank you, wassalamualaikum warahmatullahi wabarakatuh