Calculate the standard free energy change for the reaction Given that the standard free energies of formation for are -16.8, +86.7 and -237.2 kJ mol -1 respectively. Predict the feasibility of the above reaction at the standard state. - Wired Faculty

Question

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Calculate the standard free energy change for the reaction
$4 NH subscript 3 left parenthesis straight g right parenthesis space plus space 5 straight O subscript 2 space space space rightwards arrow space space space 4 NO left parenthesis straight g right parenthesis space plus space 6 straight H subscript 2 straight O left parenthesis straight l right parenthesis$

Given that the standard free energies of formation $increment subscript straight f straight G to the power of 0$ for $NH subscript 3 left parenthesis straight g right parenthesis comma space space NO left parenthesis straight g right parenthesis space and space straight H subscript 2 straight O left parenthesis straight l right parenthesis$ are -16.8, +86.7 and -237.2 kJ mol^-1 respectively. Predict the feasibility of the above reaction at the standard state.

Solution

We know,
$increment subscript straight r straight G to the power of 0 space equals space sum from blank to blank of increment subscript straight f straight G to the power of 0 left parenthesis products right parenthesis space minus space sum from blank to blank of increment subscript straight f straight G to the power of 0 left parenthesis reactants right parenthesis space space space space space space space equals open square brackets 4 space cross times space increment subscript straight f straight G to the power of 0 left parenthesis NO right parenthesis space plus space 6 cross times space increment subscript straight f straight G to the power of 0 left parenthesis straight H subscript 2 straight O right parenthesis close square brackets space space space space space space space space space$
$negative open square brackets 4 space cross times space increment subscript straight f straight G to the power of 0 left parenthesis NH subscript 3 right parenthesis space plus 5 space cross times space increment subscript straight f straight G to the power of 0 left parenthesis straight O subscript 2 right parenthesis close square brackets$
$equals open square brackets 4 space cross times space left parenthesis 86.7 right parenthesis space plus space 6 space cross times space left parenthesis negative 237.2 right parenthesis close square brackets space kJ space mol to the power of negative 1 end exponent space space space space space space space space space space space space space minus open square brackets 4 cross times left parenthesis negative 16.8 right parenthesis space plus space 5 space cross times space 0 close square brackets kJ space mol to the power of negative 1 end exponent space space space equals space open square brackets 346.8 space minus space 1423.2 close square brackets space kJ space mol to the power of negative 1 end exponent space plus space 67.2 space kJ thin space mol to the power of negative 1 end exponent space space space equals space minus 1076.4 space kJ space mol to the power of negative 1 end exponent space plus space 67.2 space kJ space mol to the power of negative 1 end exponent space space space equals negative 1009.2 space kJ$
Since $increment subscript straight r straight G to the power of 0$ is negative, the process is feasible.

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