Question

Consider the following date
Heat of combustion of H₂(g)           = -241.8 kJ mol^-1 
Heat of combustion of C(s)              = -393.5 kJ mol^-1 
Heat of combustion of C₂H₅OH(1)  = -1234.7 kJ mol^-1
The heat of formation of C₂H₅OH(1) is (-) ____ kJ mol^-1 (Nearest integer)

Answer 1

Correct Answer: 278

Hess's Law Enthalpy Calculation 

Given Reactions and Enthalpy Changes:

\( 2\text{C(s)} + \text{O}_2 \rightarrow 2\text{CO}_2 \qquad \Delta H = -393.5 \times 2 = -787 \text{ kJ} \qquad (1) \)

\( 3\text{H}_2 + \frac{3}{2}\text{O}_2 \rightarrow 3\text{H}_2\text{O} \qquad \Delta H = -241.5 \times 3 = -725.4 \text{ kJ} \qquad (2) \)

\( \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \qquad \Delta H = -1234.7 \text{ kJ} \qquad (3) \)

\( 3\text{H}_2\text{O} + 2\text{CO}_2 \rightarrow \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \qquad \Delta H = +1234.7 \text{ kJ} \qquad (4) \)

Target Reaction:

\( 2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH} \qquad (5) \)

Calculation:

Equation (5) is derived as:

\( (5) = (1) + (2) + (4) \)

\( \Delta H = (-787) + (-725.4) + (1234.7) \)

\( \Delta H = -277.7 \text{ kJ} \implies \Delta H \approx -278 \text{ kJ} \)

Answer 2

We can use Hess's Law and combine the equations to calculate the heat of formation of C\(_2\)H\(_5\)OH(l):
\( \Delta H_{\text{f}} \) = \( -393.5 \times 2 \) - \( 241.5 \times 8 \times 3 \) + 1234.7 = -277.7 kJ mol\(^{-1}\)