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Question

In a fuel cell (the device used for producing electricity directly from a chemical reaction) methanol is used as a fuel and oxygen gas is used as an oxidizer. The standard enthalpy of combustion of methanol is $$-726kJmol^{-1}$$. The standard free energies of formation of $$CH_3OH(l).CO_{2(g)}$$ and  $$H_2O (l)$$ are $$-166.3, -394.4$$ and $$-237.1 kJ mol^{-1}$$ respectively.
The standard free energy change of the reaction will be:


A
597.8kJmol1
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B
298.9kJmol1
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C
465.2kJmol1
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D
702.3kJmol1
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Solution

The correct option is C $$-465.2 kJ mol^{-1}$$
Solution:- (C) $$-465.2 \; {kJ}/{mol}$$
$${C{H}_{3}OH}_{\left( l \right)} + \cfrac{3}{2} {{O}_{2}}_{\left( g \right)} \longrightarrow {C{O}_{2}}_{\left( g \right)} + 2 {{H}_{2}O}_{\left( l \right)}$$
Given:-
$${\Delta{{{G}^{0}}_{f}}}_{\left({C{H}_{3}OH}_{\left( l \right)} \right)} = -166.3 \; {kJ}/{mol}$$
$${\Delta{{{G}^{0}}_{f}}}_{\left({C{O}_{2}}_{\left( l \right)} \right)} = -394.4 \; {kJ}/{mol}$$
$${\Delta{{{G}^{0}}_{f}}}_{\left({{H}_{2}O}_{\left( l \right)} \right)} = -237.1 \; {kJ}/{mol}$$
Therefore,
$${\Delta{{G}^{0}}}_{Reaction} = {\sum{\Delta{{{G}^{0}}_{f}}}}_{\left( product \right)} - {\sum{\Delta{{{G}^{0}}_{f}}}}_{\left( reactant \right)}$$
$$\Rightarrow {\Delta{{G}^{0}}}_{Reaction} = \left( {\Delta{{{G}^{0}}_{f}}}_{\left({C{O}_{2}}_{\left( l \right)} \right)} + {\Delta{{{G}^{0}}_{f}}}_{\left({{H}_{2}O}_{\left( l \right)} \right)} \right) - {\Delta{{{G}^{0}}_{f}}}_{\left({C{H}_{3}OH}_{\left( l \right)} \right)}$$
$$\Rightarrow {\Delta{{G}^{0}}}_{Reaction} = \left( \left( -394.4 \right) + \left( -237.1 \right) \right) - \left( -166.3 \right)$$
$$\Rightarrow {\Delta{{G}^{0}}}_{Reaction} = -465.2 \; {kJ}/{mol}$$
Hence the standard free energy change of the reaction will be $$-465.2 \; {kJ}/{mol}$$.

Chemistry

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