The Gibbs free energy of formation ((Δ )f G ° ) for the reaction: 2Mg(s)+O2(g) arrow 2MgO(s) is measured and plotted as function of temperature (T) . Of the 4 possible graph below, which one would most likely represent the observed trend?

Question

The Gibbs free energy of formation ((Δ )f G ° ) for the reaction: 2Mg(s)+O2(g) arrow 2MgO(s) is measured and plotted as function of temperature (T) . Of the 4 possible graph below, which one would most likely represent the observed trend? <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-len8b4tlltfwmy2u.jpg" /> (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

For reaction: 2Mg(s)+O2(g) arrow 2MgO(s) Δ fH=-ve Δ fS=-ve Δ fG° =Δ fH° -TΔfS° As T increase, triangle fG° increase

Q. The Gibbs free energy of formation $((Δ)_{f} G^{\circ})$ for the reaction: $2 M g (s) + O_{2} (g) \to 2 M g O (s)$ is measured and plotted as function of temperature $(T)$ . Of the $4$ possible graph below, which one would most likely represent the observed trend?

$1$

$2$

$3$

$4$

For reaction: $2 M g (s) + O_{2} (g) \to 2 M g O (s)$
$Δ_{f} H = - v e$
$Δ_{f} S = - v e$
$Δ_{f} G^{^{\circ}} = Δ_{f} H^{^{\circ}} - T Δ_{f} S^{^{\circ}}$
As $T$ increase, $△_{f} G^{\circ}$ increase

Q. The Gibbs free energy of formation ((Δ)f​G∘) for the reaction: 2Mg(s)+O2​(g)→2MgO(s) is measured and plotted as function of temperature (T) . Of the 4 possible graph below, which one would most likely represent the observed trend?

1

2

3

4

For reaction: 2Mg(s)+O2​(g)→2MgO(s) Δf​H=−ve Δf​S=−ve Δf​G∘=Δf​H∘−TΔf​S∘ As T increase, △f​G∘ increase

Q. The Gibbs free energy of formation $((Δ)_{f} G^{\circ})$ for the reaction: $2 M g (s) + O_{2} (g) \to 2 M g O (s)$ is measured and plotted as function of temperature $(T)$ . Of the $4$ possible graph below, which one would most likely represent the observed trend?

$1$

$2$

$3$

$4$

For reaction: $2 M g (s) + O_{2} (g) \to 2 M g O (s)$
$Δ_{f} H = - v e$
$Δ_{f} S = - v e$
$Δ_{f} G^{^{\circ}} = Δ_{f} H^{^{\circ}} - T Δ_{f} S^{^{\circ}}$
As $T$ increase, $△_{f} G^{\circ}$ increase