60) Calculate the standard enthalpy change for the reaction given that Ha--1669.

Question

60) Calculate the standard enthalpy change for the reaction given that Ha--1669.8 J/mol H--822 2/mol 61) Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water: Use the following information to calculate He, in kilojoules for the combustion of methane: CHdg) +202(g) COtg) +2 1120() 62) Reaction ofgaseous fluorine (F) with compound CIF yields a product of C15. Using the following information to calculate Iran for the synthesis orch 2CIF(g) +O(g)Cl0(g) +OF (g) H.. + 205.4 kJ OF:(g) AH-+24.5 kJ/mol 63) Calculate Hor in kl mol for benzene, C6H6. from the following data: 2 CaHe1)+15 O(g)> 12 CO(g)+6 H2O) AHn6534 kJ AH' (CO2)393.5 kJ/mol AH H:0)-285.8 kJ/mol 64) Water gas is the name for the mixture of CO and H: prepared by reaction of steam with carbon at 1000 Cc as) + H20(g) CO (g) + H2(g) information to calculate AHun in kilojoules for the water-gas reaction: C(s) + O2(g) CO2(g) AH--393.5 kJ 65) Given the thermochemical equations Br2(1) + F2(g) 2BrF(g) AH188 kJ/mol calculate the //oran for the reaction H768 k/mol

Explanation / Answer

Solution:- All these problems are similar and based on Hess's law. We arrange the given equations in a way so that we could get the desired equation and then add their delta H values accordingly.

(61) We need to find out the delta H for the reaction, CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)

First reactant of our equation is CH4(g) and it is present in first given equation, so let's write this..

CH4(g) + O2(g) ----> CH2O(g) + H2O(g) delta H0 = -275.6 kJ

second reactant of our equation is O2(g) and we need two moles of it. First equation we have written so far has only one mole of O2(g), so need to use the second equation as it also has one mole of O2(g).

CH2O(g) + O2(g) ----> CO2(g) + H2O(g) delta H0 = -526.7 kJ

to cancel H2O(g) and get 2H2O(l) on product side we need to flip and multiply the third given equation.

2H2O(g) ------> 2H2O(l) delta H0 = -88.0 kJ

On adding these three equations..

CH4(g) + O2(g) ----> CH2O(g) + H2O(g) delta H0 = -275.6 kJ

CH2O(g) + O2(g) ----> CO2(g) + H2O(g) delta H0 = -526.7 kJ

2H2O(g) ------> 2H2O(l) delta H0 = -88.0 kJ

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CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l) delta H0 = -890.3 kJ

(62)

We are asked to calculate the delta H0 for the reaction, F2(g) + ClF(g) ----> ClF3(g)

Standard enthalpy of formation is given for the formation of OF2(g). Standard enthalpy of formation means the formation from it's elements. The equation would be...

1/2O2(g) + F2(g) -----> OF2(g) delta H0 = 24.5 kJ

For our equation we need F2(g) and the equation we have just written as this so let's move to the next reactant we want. next we need ClF(g) which is present in first equation. First equation has 2 moles of it so we need to divide it by 2.

ClF(g) + 1/2O2(g) -----> 1/2Cl2O(g) + 1/2OF2(g) delta H0 = 102.7 kJ

Now we want one ClF3(g) on product side. second equation has 2 moles of it on reactant side so we nede to flip it and divide by 2.

1/2ClO2(g) + 3/2OF2(g) ----> ClF3(g) + O2(g) delta H0 = -266.4 kJ

let's add these three equations..

1/2O2(g) + F2(g) -----> OF2(g) delta H0 = 24.5 kJ

ClF(g) + 1/2O2(g) -----> 1/2Cl2O(g) + 1/2OF2(g) delta H0 = 102.7 kJ

1/2ClO2(g) + 3/2OF2(g) ----> ClF3(g) + O2(g) delta H0 = -266.4 kJ

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F2(g) + ClF(g) ------> ClF3(g) delta H0 = -139.2 kJ

(63)

delta H0 of reaction = sum enthalpy of formation of products - sum of enthalpy of formation of reactants

delta H0 of reaction = [12(CO2) + 6(H2O)] - [2(C6H6) + 15(O2)]

-6534 = [(12(-393.5) + 6(-285.8)] - [2(X) + 15(0)]

-6534 = -6436.8 - 2X

2X = -6436.8 + 6534

2X = 97.2

X = 97.2/2 = 48.6 kJ/mol

So, the standard enthalpy of formation of benzene is +48.6 kJ/mol.

(65)

Need to divide the first equation and also flip it to get BrF(g) on reactant side.

BrF(g) -----> 1/2Br2(g) + 1/2F2(g) delta H0 = 94 kJ

to get one mole of BrF3(g) we nede to divide the second equation by 2.

1/2Br2(g) + 3/2F2(g) -----> BrF3(g) delta H = -384 kJ

on adding these two equations....

BrF(g) -----> 1/2Br2(g) + 1/2F2(g) delta H0 = 94 kJ

1/2Br2(g) + 3/2F2(g) -----> BrF3(g) delta H = -384 kJ

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BrF(g) + F2(g) -----> BrF3(g) delta H0 = -290 kJ

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