Question

1. Consider the following reaction:

             C₂H₂ (g)+ 52 O₂ (g) ® 2 CO₂ (g) + H₂O (g)                      

1. Given ΔHf° of CO₂ (g) = -393.5 KJ/mol, ΔHf° H₂O (g) = -241.8 KJ/mol, and ΔHf° for

C₂H₂ (g) = 227.4 KJ/mol, calculate ΔHrxn° for this reaction.

How many KJ of heat is released when 0.440 kg of carbon dioxide produced?

Answer 1

a)

Given:
Hof(C2H2(g)) = 227.4 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol
Hof(CO2(g)) = -393.5 KJ/mol
Hof(H2O(g)) = -241.8 KJ/mol

Balanced chemical equation is:
C2H2(g) + 5/2 O2(g) ---> 2 CO2(g) + H2O(g)

ΔHo rxn = 2*Hof(CO2(g)) + 1*Hof(H2O(g)) - 1*Hof( C2H2(g)) - 5/2*Hof(O2(g))
ΔHo rxn = 2*(-393.5) + 1*(-241.8) - 1*(227.4) - 5/2*(0.0)
ΔHo rxn = -1256.2 KJ
Answer: -1256.2 KJ

b)

Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol


mass(CO2)= 0.440 Kg
= 440.0 g

use:
number of mol of CO2,
n = mass of CO2/molar mass of CO2
=(4.4*10^2 g)/(44.01 g/mol)
= 9.998 mol
Since Δ H is negative, heat is released
when 2 mol of CO2 reacts, heat released = 1256.2 KJ
So,
for 9.998 mol of CO2, heat released = 9.998*1256.2/2 KJ
= 6.28*10^3 KJ
Answer: 6.28*10^3 KJ