Given that :2 C_{(s)} + 2 O_{2_{(g)}} rightar | vedclass

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(D) The formation reaction for acetylene is $:$$2 C_{(s)} + H_{2_{(g)}} \rightarrow C_2 H_{2_{(g)}}$Using the given enthalpies of combustion $:$$\Delt

Vedclass · Accepted Answer

$ +228 \ kJ$

Vedclass · Answer

(D) The formation reaction for acetylene is $:$$2 C_{(s)} + H_{2_{(g)}} ightarrow C_2 H_{2_{(g)}}$Using the given enthalpies of combustion $:$$\Delta_{f} H^{\circ} = [2 \Delta_{c} H^{\circ}(C) + \Delta_{c} H^{\circ}(H_2)] - \Delta_{c} H^{\circ}(C_2 H_2)$Note that the first equation given is for $2 \ mol$ of $C$,so $\Delta_{c} H^{\circ}(C) = \frac{-787}{2} \ kJ/mol$.$\Delta_{f} H^{\circ} = [2 	imes (\frac{-787}{2}) + (-286)] - (-1301)$$\Delta_{f} H^{\circ} = [-787 - 286] + 1301$$\Delta_{f} H^{\circ} = -1073 + 1301 = +228 \ kJ/mol$.

Similar Questions

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ Given that: $\Delta_{f} H^{\circ}(CH_4) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$

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