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In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is

CH_{3}OH_{\left ( l \right )}+\frac{3}{2}O_{2\left ( g \right )}\rightarrow CO_{2\left ( g \right )}+2H_{2}O_{\left ( l \right )}

At 298 K standard Gibb’s energies of formation for CH_{3}OH_{\left ( l \right )},H_{2}O_{\left ( l \right )}\: and\: CO_{2\left ( g \right )}

are –166.2, –237.2 and –394.4 kJ mol-1 respectively. If standard enthalpy of combustion of methanol is –726 kJ mol-1 , efficiency of the fuel cell will be

  • Option 1)

    80

  • Option 2)

    87

  • Option 3)

    90

  • Option 4)

    97

 

Answers (1)

best_answer

Concept missing

Efficiency of fuel cell (\eta)=\frac{\Delta G^{o}}{\Delta H^{o}}=\frac{-702.6}{1-726}\times 100=97 \ ^{o}/_{o}


Option 1)

80

Incorrect

Option 2)

87

Incorrect

Option 3)

90

Incorrect

Option 4)

97

Correct

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Plabita

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