JEE Advance - Chemistry (2015 - Paper 1 Offline - No. 3)
All the energy released from the reaction
$$X \to Y, \Delta _tG^o $$ = -193 kJ mol-1 is used for oxidizing M+ as M+ $$\to$$ M3+ + 2e-, Eo = -0.25 V
Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1]
$$X \to Y, \Delta _tG^o $$ = -193 kJ mol-1 is used for oxidizing M+ as M+ $$\to$$ M3+ + 2e-, Eo = -0.25 V
Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1]
Answer
4
Explanation
Given :
X $$\to$$ Y; $$\Delta$$rG$$^\circ$$ = $$-$$ 193 kJ mol$$-$$1
M+ $$\to$$ M3+ + 2e$$-$$; E$$^\circ$$ = $$-$$0.25 V
F = 96500 C mol$$-$$1
Let 193 kJ is used for oxidising x moles of M+.
For 1 mole of M+,
$$\Delta$$G$$^\circ$$ = $$-$$nFE$$^\circ$$
= $$-$$2 $$\times$$ 96500 $$\times$$ ($$-$$0.25)
= 48250 J mol$$-$$1 = 48.25 kJ mol$$-$$1
Thus, no. of moles of M+ oxidized when one mole of X is converted to Y = $${{193} \over {48.25}} = 4$$.
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