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

Question

In which of the following reactions, standard entropy change (Δ S°) is positive and standard Gibb’s energy change (Δ G°) decreases sharply with increasing temperature ? (A) C (graphite) +(1/2) O 2( g ) longrightarrow CO ( g ) (B) CO ( g )+(1/2) O 2( g ) longrightarrow CO 2( g ) (C) Mg ( s )+(1/2) O 2( g ) longrightarrow Mg O ( s ) (D) (1/-2) C (graphite) +(1/2) O 2( g ) longrightarrow (1/2) CO 2( g )

Tardigrade Admin · Accepted Answer

C (graphite) +(1/2) O 2( g ) arrow CO ( g ) Δ S ° is positive for this reaction because 0.5 mole of gas on reactant side is giving one mole of gas on product side after reacting with C (graphite). Also Δ G decreases sharply with increase in temperature.

Q. In which of the following reactions, standard entropy change $(Δ S^{\circ})$ is positive and standard Gibb’s energy change $(Δ G^{\circ})$ decreases sharply with increasing temperature ?

$C$ (graphite) $+ \frac{1}{2} O_{2} (g) ⟶ CO (g)$

$CO (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)$

$M g (s) + \frac{1}{2} O_{2} (g) ⟶ M g O (s)$

$\frac{1}{- 2} C$ (graphite) $+ \frac{1}{2} O_{2} (g) ⟶ \frac{1}{2} C O_{2} (g)$

$C$ (graphite) $+ \frac{1}{2} O_{2} (g) \to CO (g)$
$Δ S^{\circ}$ is positive for this reaction because $0.5$ mole of gas on reactant side is giving one mole of gas on product side after reacting with $C$ (graphite).
Also $Δ G$ decreases sharply with increase in temperature.

Q. In which of the following reactions, standard entropy change (ΔS∘) is positive and standard Gibb’s energy change (ΔG∘) decreases sharply with increasing temperature ?

C (graphite) +21​O2​(g)⟶CO(g)

CO(g)+21​O2​(g)⟶CO2​(g)

Mg(s)+21​O2​(g)⟶MgO(s)

−21​C (graphite) +21​O2​(g)⟶21​CO2​(g)