Question

At 25°C and 1 atm pressure, the enthalpy of combustion of benzene (I) and acetylene (g) are –3268 kJ mol–1 and –1300 kJ mol–1, respectively. The change in enthalpy for the reaction 3C2H2(g) → C6H6(I), is

• A. +324 kJ mol–1
• B. +632 kJ mol–1
• C. –632 kJ mol–1
• D. –732 kJ mol–1

Answer 1

Answer: (A)

+324 kJ mol–1

Answer 2

The correct option is(C): –632 kJ mol–1.

I.

$C_6H_6(l)+\frac{15}{2}O_2(g)→6CO_2(g)+3H_2O(g)$

ΔH1 = –3268 kJ/mol

II.

$C_2H_2(g)+\frac{5}{2}O_2(g)→2CO_2(g)+H_2O(g)$

ΔH2 = –1300 kJ/mol

III.

$3C_2H_2(g)-2C→6H_6(l)$

ΔH3

Applying Hess’s law of constant heat summation

ΔH3 = 3 × ΔH2 – ΔH1

= 3 × (–1300) – (–3268)

= –632 kJ/mol