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Question

(C) The enthalpy of reaction $(\Delta_r H^0)$ is calculated using the formula:$\Delta_r H^0 = \sum \Delta_f H^0 (	ext{products}) - \sum \Delta_f H^0

Vedclass · Accepted Answer

$-1390$

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(C) The enthalpy of reaction $(\Delta_r H^0)$ is calculated using the formula:$\Delta_r H^0 = \sum \Delta_f H^0 (	ext{products}) - \sum \Delta_f H^0 (	ext{reactants})$For the reaction: $C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} ightarrow 2CO_{2(g)} + 3H_2O_{(l)}$$\Delta_r H^0 = [2 	imes \Delta_f H^0 (CO_2) + 3 	imes \Delta_f H^0 (H_2O)] - [\Delta_f H^0 (C_2H_5OH) + \frac{7}{2} \Delta_f H^0 (O_2)]$Given: $\Delta_f H^0 (O_2) = 0 \ kJ \ mol^{-1}$ (standard state element).$\Delta_r H^0 = [2(-400) + 3(-290)] - [-280 + 0]$$\Delta_r H^0 = [-800 - 870] + 280$$\Delta_r H^0 = -1670 + 280 = -1390 \ kJ \ mol^{-1}$

Molecule	$\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$	$-280$
$CO_{2(g)}$	$-400$
$H_2O_{(l)}$	$-290$

Calculate $\Delta_r H$ $(kJ \ mol^{-1})$ of the following reaction:
$C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$
Molecule $\Delta_f H^0 (kJ \ mol^{-1})$
$C_2H_5OH_{(l)}$ $-280$
$CO_{2(g)}$ $-400$
$H_2O_{(l)}$ $-290$

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Calculate $\Delta_r H$ $(kJ \ mol^{-1})$ of the following reaction:$C_2H_5OH_{(l)} + \frac{7}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)}$Molecule$\Delta_f H^0 (kJ \ mol^{-1})$$C_2H_5OH_{(l)}$$-280$$CO_{2(g)}$$-400$$H_2O_{(l)}$$-290$