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When combustion of Methane ( $CH_{4}$ ) takes place it releases carbon dioxide and water. If enthalpy of formation of $CH_{4}$ is $10J$ , for $O_{2}$ is $0J$ , for $CO_{2}$ is $5J$ and for $H_{2}O$ is $3J$ and only $50$ % $CH_{4}$ is converted then enthalpy of reaction and nature of reaction is:
Option: 1
$-0.5 J$ , Exothermic

Option: 2
$+0.5 J$ , Endothermic

Option: 3
$-1\, J$ , Exothermic

Option: 4
$+1\, J$ , Endothermic

Answers (1)

best_answer

Enthalpy of reaction , $\Delta H=H_{P}-H_{R}$

First write the balanced equation of the reaction

$CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O$

Given

$\Delta H_{f}^{0}CH_{4}=10J$

$\Delta H_{f}^{0}O_{2}=0J$

$\Delta H_{f}^{0}CO_{2}=5J$

$\Delta H_{f}^{0}H_{2}O=3J$

As the conversion is only $50$ % then all reactants and products will also be converted $50$ %.

So, $\Delta H=$ $5\times0.5 + 2\times3\times0.5 - 0 -10\times0.5 = 2.5+3-5 = +0.5 J$

As $\Delta H> 0,$ hence it is an endothermic process

Hence, Option number (2) is correct

Posted by

Divya Prakash Singh

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