The reaction in which standard reaction entropy change (.(ΔS)o.) is positive and standard Gibb's energy change (.(ΔG)o.) decreases sharply with increasing temperature is:

Question

The reaction in which standard reaction entropy change (.(ΔS)o.) is positive and standard Gibb's energy change (.(ΔG)o.) decreases sharply with increasing temperature is: (A) C (graphite) +(1/2)(O2)(. g .) overset arrow (CO)(. g .) (B) (CO)g+(1/2)(O2)(. g .) overset arrow (CO)(. g .) (C) (Mg)s+(1/2)(O2)(. g .) overset arrow (1/2)((CO)2)(. g .) (D) (1/2)C(graphite)+(1/2)(O2)(. g .) overset arrow (1/2)((CO)2)(. g .)

Tardigrade Admin · Accepted Answer

Cg r a p h i t e+(1/2)(O2)(. g .) arrow COg Δ ng=1-(1/2)=(1/2) As amount of gaseous substance is increasing in product, thus Δ S is positive for this reaction. And we know that Δ G=Δ H-TΔ S As Δ S is positive, thus increase in temperature will make TΔ S more negative and Δ G will decrease.

Q. The reaction in which standard reaction entropy change $((Δ S)^{o})$ is positive and standard Gibb's energy change $((Δ G)^{o})$ decreases sharply with increasing temperature is:

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

$(CO)_{g} + \frac{1}{2} (O_{2})_{(g)} \to (CO)_{(g)}$

$(M g)_{s} + \frac{1}{2} (O_{2})_{(g)} \to \frac{1}{2} ((CO)_{2})_{(g)}$

$\frac{1}{2} C (g r a p hi t e) + \frac{1}{2} (O_{2})_{(g)} \to \frac{1}{2} ((CO)_{2})_{(g)}$

Q. The reaction in which standard reaction entropy change ((ΔS)o) is positive and standard Gibb's energy change ((ΔG)o) decreases sharply with increasing temperature is:

C (graphite) +21​(O2​)(g)​→(CO)(g)​

(CO)g​+21​(O2​)(g)​→(CO)(g)​

(Mg)s​+21​(O2​)(g)​→21​((CO)2​)(g)​

21​C(graphite)+21​(O2​)(g)​→21​((CO)2​)(g)​

Cgraphite​+21​(O2​)(g)​→COg​ Δng​=1−21​=21​ As amount of gaseous substance is increasing in product, thus ΔS is positive for this reaction. And we know that ΔG=ΔH−TΔS As ΔS is positive, thus increase in temperature will make TΔS more negative and ΔG will decrease.

Q. The reaction in which standard reaction entropy change $((Δ S)^{o})$ is positive and standard Gibb's energy change $((Δ G)^{o})$ decreases sharply with increasing temperature is:

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

$(CO)_{g} + \frac{1}{2} (O_{2})_{(g)} \to (CO)_{(g)}$

$(M g)_{s} + \frac{1}{2} (O_{2})_{(g)} \to \frac{1}{2} ((CO)_{2})_{(g)}$

$\frac{1}{2} C (g r a p hi t e) + \frac{1}{2} (O_{2})_{(g)} \to \frac{1}{2} ((CO)_{2})_{(g)}$