Question

Part A; Calculate the standard enthalpy change for the reaction

2A+B⇌2C+2D

where the heats of formation are given in the following table:

Substance	ΔH∘f (kJ/mol)
A	-251
B	-375
C	191
D	-519

Answer to Part A- 221kJ  

FIND PART B & C

PART B- For the reaction given in Part A, how much heat is absorbed when 3.40 mol of A reacts?

Express your answer numerically in kilojoules.

PART C- For the reaction given in Part A, ΔS∘rxn is 29.0 J/K . What is the standard Gibbs free energy of the reaction, ΔG∘rxn?

Express your answer numerically in kilojoules.

Many thanks!!!

Answer 1

Substance..ΔH∘f
(kJ/mol)
A.......-251
B......   -375
C... 191
D.... -519

use these numbers to calculate dHrxn of the given reaction
2A+B⇌2C+2D

dHrxn = dH{product}-dH{reactants}
= 2*dH(C) + 2* dH(D) -{ 2*dH(A) + 1*dH(B)}

plug in dH values from the table

dHrxn = {2* (191) +2*(-519)}KJ - { 2(-251) +1* (-375) }KJ
= 221 KJ
***************

the above dHxn is for 2 mole of A ,as per given equatiion
So for 1 mole of A : dHrxn = 221 kj/ 2 = 110.5 KJ

So when we use 3.4 mol of A :
dHrxn = 110.5 KJ* 3.4 mol = 375.7 KJ
***************
given
dS0rxn = 29.0 J/K
dHrxn = 221x10^3 J
standard gibbs energy is calculated using formula
dG0 = dHrxn-TdSrxn

at standard temperature T= 298 K

dG0 = 221 x10^3 j -298 k * 29.0 J/K
= 212.358 KJ
******************************
All solved ,kindly upvote