In which of the following reactions,standard | vedclass

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(A) For a reaction,$\Delta G^o = \Delta H^o - T \Delta S^o$. If $\Delta S^o > 0$,then as temperature $T$ increases,the term $-T \Delta S^o$ becomes mo

Vedclass · Accepted Answer

$C_{(graphite)} + \frac{1}{2} O_{2(g)} 	o CO_{(g)}$

Vedclass · Answer

(A) For a reaction,$\Delta G^o = \Delta H^o - T \Delta S^o$. If $\Delta S^o > 0$,then as temperature $T$ increases,the term $-T \Delta S^o$ becomes more negative,causing $\Delta G^o$ to decrease sharply.$\Delta S^o$ is positive when the number of moles of gaseous products is greater than the number of moles of gaseous reactants (i.e.,$\Delta n_g > 0$).In option $A$: $C_{(graphite)} + \frac{1}{2} O_{2(g)} 	o CO_{(g)}$,$\Delta n_g = 1 - 0.5 = 0.5 > 0$.In option $B$: $CO_{(g)} + \frac{1}{2} O_{2(g)} 	o CO_{2(g)}$,$\Delta n_g = 1 - 1.5 = -0.5 In option $C$: $Mg_{(s)} + \frac{1}{2} O_{2(g)} 	o MgO_{(s)}$,$\Delta n_g = 0 - 0.5 = -0.5 In option $D$: $\frac{1}{2} C_{(graphite)} + \frac{1}{2} O_{2(g)} 	o \frac{1}{2} CO_{2(g)}$,$\Delta n_g = 0.5 - 0.5 = 0$.Thus,only reaction $A$ has a positive $\Delta S^o$.

In which of the following reactions,standard reaction entropy change $(\Delta S^o)$ is positive and standard Gibb's energy change $(\Delta G^o)$ decreases sharply with increasing temperature?

Similar Questions

Statement-$1$: For a process at equilibrium,the total entropy is maximum and the change in total entropy is zero.
Statement-$2$: For a reversible process,if the same amount of heat is supplied to a system at two different temperatures,then the entropy change is less at a lower temperature.

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