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

Correct answer: A. C (graphite) +(1)/(2) O_2(g) CO(g) Among the given reactions only in the case of C (graphite) +(1)/(2) O_2( g) CO(g) entropy increases because randomness (disorder) increases. Thus, standard entropy change (Δ S^ ) is positive. Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e., Δ H^ =- ve We know that aligned & Δ G^ =Δ H^ -T Δ S^ & Δ G^ =-v v-T(+ve) aligned Thus, as the temperature increases, the value of Δ G^ decreases.

NEET 2012

Exam: NEET
Year: 2012
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2012 · Chemistry · Thermodynamics (C)

In which of the following reactions, standard reaction entropy...

In which of the following reactions, standard reaction entropy changes

\left(\Delta S^{\circ}\right)

is positive and standard Gibb's energy change

\left(\Delta G^{\circ}\right)

decreases sharply with increasing temperature?

Options

A $\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g)$
B $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}_2(g)$
C $\mathrm{Mg}(s)+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{MgO}(s)$
D $\frac{1}{2} \mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \frac{1}{2} \mathrm{CO}_2(g)$

Correct answer

A. $\mathrm{C}$ (graphite) $+\frac{1}{2} \mathrm{O}_2(g) \longrightarrow \mathrm{CO}(g)$

Step-by-step solution

Among the given reactions only in the case of

\mathrm{C} \text{ (graphite) }+\frac{1}{2} \mathrm{O}_2(\mathrm{ g}) \longrightarrow \mathrm{CO}(\mathrm{g})

entropy increases because randomness (disorder) increases. Thus, standard entropy change

\left(\Delta S^{\circ}\right)

is positive. Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e.,

\Delta H^{\circ}=-

ve We know that

\begin{aligned} & \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\ & \Delta G^{\circ}=-v \mathrm{v}-T(+\mathrm{ve}) \end{aligned}

Thus, as the temperature increases, the value of

\Delta G^{\circ}

decreases.

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