hes law I will teach you the complete concept of hes law and fast method to calculate numerical problems of hess's law in less than a minute firstly let me teach you one daily life example let consider this ice cube it absorb DH is equal to 300 KJ energy from the surrounding as a result it sublimes to gaseous state this staric path are one step process now consider that this ice absorb dh1 is equal to 44 K energy and melt downs to liquid our water lastly this liquid absorb dh2 is equal to 256 K energy and evaporate to gaseous State now listen carefully I needs DH is equal to 300 KJ energy to convert to gas directly secondly if we add dh1 plus dh2 it will be equal to DH we know that DH is equal to 300 K while dh1 is equal to 44 K and dh2 is equal to 256 K I get 3 300 K is equal to 300 K this left hand side is equal to right hand side therefore we say that enthalpy change is the same whether it takes place in one step are multiple steps let me repeat it enthalpy change is the same whether it takes in one step or multiple steps this concept is called hes law which we will learn in the next slide hence noted down this practical example of H's law now what is H's law well consider this reactant a let it absorb DH is equal to 24 KJ energy and convert to product D let this whole reaction occur in one step now let consider that this reaction takes place in multiple steps to form product let this air reactant absorb dh1 is equal to 12 K and convert to B this B absorb dh2 is equal to 8 K and convert to C and this C absorb dh3 is equal to 4 K and convert to D here H Baba states that DH is equal to dh1 plus dh2 plus dh3 he means that in the first step DH is equal to 24 K dh1 is equal to 12 K dh2 is equal to 8 K and dh3 is equal to 4 K after calculation I get 24 K is equal to 24 K this left hand side is equal to right hand side therefore H Baba states that enthalpy change during a chemical reaction is the same whether it takes place in one or more than one steps let me repeat it enthalpy change during a chemical reaction is the same whether it takes place in one or more than one steps this is the first statement our statement number one of H law now we will learn the second statement of the H's law we know that DH is equal to dh1 plus dh2 plus dh3 here I shift all these terms from the right hand side to the left hand side I write DH minus dh1 minus dh2 minus dh3 we know that DH is equal to 24 KJ while dh1 dh2 and dh3 are also equal to 24 K 24 Kus - 24 K is equal to 0 K it means that total energy in this cycle is zero are conserved let me repeat it it means that total energy in the cycle is zero are conserved therefore H Baba states that enthalpy change and ayic process is always zero or conserved mathematically we write sumission of all enthalpy change and a cyclic process is zero hence this is the second statement of H's law just remember this basic concept of H's law now we will learn important example of H's law let consider the formation of carbon dioxide gas in one state carbon in a solid state plus oxygen gas react together to form carbon dioxide gas remember that the physical state of each ingredient is very important in this type of reactions the enthalpy change for this reaction is DH is equal to 393.7 K per mole here this minus sign means that the reaction is exothermic because heat is given off now we will learn the formation of carbon dioxide oxide gas in two stepes in the first St carbon plus half of oxygen gas react together to form carbon monooxide gas the enthalpy change dh1 is equal to minus 111052 K per mole in the Second Step carbon monooxide gas plus half of oxygen gas react together to form carbon dioxide gas is the enthalpy change for this reaction dh2 is equal to - 393 K per mole now I will add these two reactions we can see that this carbon monooxide and this carbon monoxide cancel out I write this carbon half of this oxygen gas and this oxygen gas becomes oxygen gas I write this carbon dioxide gas here according to H's law DH is equal to dh1 + dh2 we know that D H1 is equal to- 111052 K per mole while dh2 is equal to 283 K per mole after calculation I get DH is equal to - 39352 k per mole thus in one step DH is equal to - 393 52 K per mole while in these two stabes the D is again are still minus 3 93.5 2 K per mole therefore H Baba states that enthalpy change in a chemical reaction is the same whether it occurs in one step or multiple stepes hence noted down this important example here let me teach you one of my favorite questions why we need to study H's law well we know that we use Kor meter to measure enthalpy change of a chemical reaction but there are many reactions like formation of ccl4 whose enthalpy change cannot be measured by using calor meter in such cases is we use H's law to determine their inaly change for this reason we need to study hes law hence noted down this important question now we will learn numerical problems of H's law I will teach you the super fast method to solve any numerical problem of hesis law in less than a minute consider this ncrt numerical problem calculate the standard in enpy of formation of methanol from the following data well standard enalia of formation means when one mole of a compound is formed from its stable elements for example stable element of carbon is solid carbon stable element of hydrogen is H2 or hydrogen gase stable element of oxygen is O2 or oxygen gase here methanol is formed from stable elements it means that methanol is made from carbon plus H2 + O2 now I balance the equation there are four hydrogen atoms I write here two there is only one oxygen atom I write here 1 by two here I write the required reaction now listen carefully in the required reaction methanol ch3 is present at the right hand side while in the first reaction methanol ch3 is present at the left hand side now to bring this methanol to the right hand side as per required reaction I reverse this reaction remember that when I reverse any reaction I change the sign of enthalpy let me repeat this important point when I reverse any reaction I change the sign of enthalpy so I write here the right hand side of this reaction carbon dioxide plus 2 H2O and then I write the left hand side of this reaction methanol ch3 plus 3x2 oxygen gas the sign of enthalpy Chang from minus to positive I get dh1 is equal to + 726 K secondly in the required reaction there is one carbon present at the left hand side while in the second reaction the carbon is also present at the left hand side or at the same side so I write this reaction as it is carbon + O2 and then I write carbon dioxide the enthalpy change is DH is equal to - 393 KJ thirdly in the required reaction at the left hand side there are two molecules of hydrogen gas while in the third reaction there is one molecule of hydrogen gas at the left hand side we can see that in the required reaction there are two hydrogen gas molecules while in the third reaction there is only one hydrogen gas molecule so I multiply this reaction by two also remember that I multiply its enthalpy change by two I get two hydrogen gas molecule plus 2 n22 oxygen gas and 2 H2O now I multiply this enthalpy change by 2 2 and2 - 286 k is equal to - 572 K now I will add these three reactions I cancel all the spectator molecules this carbon dioxide and this carbon di oxide cancel out these two water molecule and these two water molecules also can cancel out this 3x2 means one full oxygen gas plus 1 half oxygen gas I cancel this full oxygen gas and half of this oxygen gas I also cancel this 3x2 oxygen gas I get carbon + 2 H2 + 1X 2 oxygen gas and this methanol ch3 thus using this fast method we can easily get the required reaction now according to hes Baba DH is equal to dh1 plus DH 2 + dh3 I write DH is equal to dh1 is + 726 minus DH H2 is - 393 and dh3 is - 572 after calculation I get - 239 KJ therefore - 239 K energy is the enthalpy ch CH for the standard formation of methanol hence noted down this fast method of calculation of enthalpy change finally consider this difficult numerical problem calculate the enthalpy change for the reaction from the following data well I use the first method I write this is the required reaction now in the required reaction Fe2O3 is present at the left hand side while in the first reaction Fe2O3 is also present at the left hand side are at the required side there is one unit of Fe2O3 and the required reaction also there is one unit in the first reaction so I write the first reaction as it is f23 + 3 carbon monooxide and then I write 2 2 Fe + 3 carbon dioxide gas the enthalpy change jh1 is equal to - 26 kilj secondly in the required reaction to feo is present at the right hand side while in the second reaction One feo is present at the left hand side to balance the two feo unit and required reaction I multiply this reaction by two also I reverse the second reaction because the required units are present at the opposite sides I write 2 Fe + 2 and 2 feo + 2 carbon monooxide I change the sign of enthalpy dh2 is equal to also I multiply it by two 2 N to + 16 K now listen carefully if I at this reaction I get something else I mean 3 carbon monoxide plus 2 carbon monoxide is equal to five carbon monooxide while in the required reaction there is only one carbon monooxide now I will try another method If I subtract three carbon monoxide minus 2 carbon monooxide I get one carbon monooxide which is the correct number of car carbon monooxide according to the required reaction I subtract reaction two from the reaction One these two Fe and these two Fe cancel out I get Fe2O3 plus 3 carbon monoxide minus 2 carbon monooxide is equal to 1 carbon monooxide I write these two F EO these three carbon dioxide minus these two carbon monooxide is equal to one carbon dioxide thus I got the required reaction now according to hpaba DH is equal to dh1 + dh2 our DH is equal to - 26 k + 32 K after calculation I get DH is equal to + 6 K hence the enthalpy change for this reaction is DH is equal to + 6 k so using this fast method we can easily crack any numerical problem of hes law I hope that you have learned all about hes law