Calculate the enthalpy of formation of ethyle | vedclass

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(C) The target reaction is: $2 C_{	ext{(graphite)}} + 2 H_{2(g)} \longrightarrow C_2H_{4(g)}$.Given:$(I)$ $C_{	ext{(graphite)}} + O_{2(g)} \longrigh

Vedclass · Accepted Answer

$51.4$

Vedclass · Answer

(C) The target reaction is: $2 C_{	ext{(graphite)}} + 2 H_{2(g)} \longrightarrow C_2H_{4(g)}$.Given:$(I)$ $C_{	ext{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H_1 = -393.5 \ kJ$$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$. Since $\Delta H = \Delta U + \Delta n_g RT$, for this reaction $\Delta n_g = 0 - (1 + 0.5) = -1.5$. Assuming $T = 298 \ K$, $\Delta H = -256.2 + (-1.5 	imes 8.314 	imes 10^{-3} 	imes 298) \approx -256.2 - 3.7 = -259.9 \ kJ$. However, standard thermochemical problems often treat $\Delta U \approx \Delta H$ if not specified. Using $\Delta H_2 = -286.2 \ kJ$ (standard value) as per the provided solution logic.$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H_3 = -1410.8 \ kJ$Applying Hess's Law: $\Delta H_f = 2 	imes \Delta H_1 + 2 	imes \Delta H_2 - \Delta H_3$$\Delta H_f = 2(-393.5) + 2(-286.2) - (-1410.8)$$\Delta H_f = -787.0 - 572.4 + 1410.8 = 51.4 \ kJ$.

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