OnlineCBT
Chemistry · THERMODYNAMICS
हिन्दीGiven 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 \( 2\text{CO}(g) \rightarrow 2\text{C}(\text{graphite}, s) + \text{O}_2(g) \)?
Correct answer · Option 1
\( +221.0\text{ kJ} \)
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 \( \text{CO} \)), \( \Delta H = -2 \times (-110.5) = +221.0\text{ kJ} \).
