Calculate the enthalpy of formation of nitric | vedclass

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(B) The reaction is: $NO_{(g)} + CO_{(g)} \rightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}$The enthalpy of reaction is given by: $\Delta H^o = [\sum \Delt

Vedclass · Accepted Answer

$89.2$

Vedclass · Answer

(B) The reaction is: $NO_{(g)} + CO_{(g)} ightarrow \frac{1}{2} N_{2(g)} + CO_{2(g)}$The enthalpy of reaction is given by: $\Delta H^o = [\sum \Delta H_f^o (	ext{products})] - [\sum \Delta H_f^o (	ext{reactants})]$$\Delta H^o = [\frac{1}{2} \Delta H_f^o (N_2) + \Delta H_f^o (CO_2)] - [\Delta H_f^o (NO) + \Delta H_f^o (CO)]$Since $\Delta H_f^o (N_2) = 0$ (standard state of an element),we have:$-372.2 = [0 + (-393.5)] - [\Delta H_f^o (NO) + (-110.5)]$$-372.2 = -393.5 - \Delta H_f^o (NO) + 110.5$$-372.2 = -283.0 - \Delta H_f^o (NO)$$\Delta H_f^o (NO) = -283.0 + 372.2 = 89.2 \, kJ \, mol^{-1}$

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