Question

Free Energy and Thermodynamics

Example #1

H₂ (g) + Cl₂ (g) à 2HCl (g) DeltaH = -185 kJ

Consider the reaction above:

Calculate Delta H

a. 1.00 mol of HCl is formed

b. 1.00 g of Cl₂ reacts

c. 2.50 L of HCl (g) at 50.0^oC and 725 mm Hg is formed.

Answer 1

a)
from reaction,
delta H = (1/2)*moles of HCl * delta H rxn
= (1/2)*1.00 mol * (-185 KJ)
= - 92.5 KJ

Answer: - 92.5 KJ

b)

Molar mass of Cl2 = 70.9 g/mol


mass(Cl2)= 1.00 g

number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(1.0 g)/(70.9 g/mol)
= 1.41*10^-2 mol

from reaction,
delta H = moles of HCl * delta H rxn
= 1.41*10^-2 mol * (-185 KJ)
= - 2.61 KJ

Answer: - 2.61 KJ

c)
Given:
P = 725.0 mm Hg
= (725.0/760) atm
= 0.9539 atm
V = 2.5 L
T = 50.0 oC
= (50.0+273) K
= 323 K

find number of moles using:
P * V = n*R*T
0.9539 atm * 2.5 L = n * 0.08206 atm.L/mol.K * 323 K
n = 8.993*10^-2 mol

from reaction,
delta H = (1/2)*moles of HCl * delta H rxn
= (1/2)*8.993*10^-2 mol * (-185 KJ)
= - 8.32 KJ

Answer: - 8.32 KJ