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Chemistry · THERMODYNAMICS

हिन्दी

Given the thermochemical equations: (i) \( \text{C}(\text{graphite}, s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \); \( \Delta_r H_1^\ominus = -393.5\text{ kJ mol}^{-1} \) (ii) \( \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \); \( \Delta_r H_2^\ominus = -283.0\text{ kJ mol}^{-1} \) What is the enthalpy change for the conversion of \( 56\text{ g} \) of \( \text{CO}(g) \) back into graphite and dioxygen according to \( 2\text{CO}(g) \rightarrow 2\text{C}(\text{graphite}, s) + \text{O}_2(g) \)?

  1. A
    \( +221.0\text{ kJ} \)
  2. B
    \( +110.5\text{ kJ} \)
  3. C
    \( -221.0\text{ kJ} \)
  4. D
    \( -110.5\text{ kJ} \)
View Answer & Solution

Correct answer · Option A

\( +221.0\text{ kJ} \)

Explanation

\( \Delta_f H^\ominus(\text{CO}) = -393.5 - (-283.0) = -110.5\text{ kJ mol}^{-1} \). For the reverse reaction of 2 moles (\( 56\text{ g} \) is \( 2\text{ mol} \) of \( \text{CO} \)), \( \Delta H = -2 \times (-110.5) = +221.0\text{ kJ} \).
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