Heat of combustion C(s),H2(g) and CH4(g) are -94 , -68 and -213 kcal/mol. Then Δ H° for C(s)+2H2(g)→Δ CH4(g)is

Question

Heat of combustion C(s),H2(g) and CH4(g) are -94 , -68 and -213 kcal/mol. Then Δ H° for C(s)+2H2(g)→Δ CH4(g)is (A) -17 kcal (B) -111 kcal (C) -170 kcal (D) -85 kcal

Tardigrade Admin · Accepted Answer

For reaction, C(s)+2H2(g)→ CH4(g), Δ H°=? Δ H°=-[(Δ H°cofCH4)-(Δ H°cof C+2×Δ H°cof H2)] C+O2→ CO2;Δ H=-94 kcal ...(i) 2H2+O2→ 2H2O;Δ H=-68×2 kcal ...(ii) CH4+2O2→ CO2+2H2O;Δ H=-213 kcal ...(ii) Eq (i)+(ii)-Eq(iii) =-[(-213)-(-94+2×-68)]kcal/mol =-[-213+230]=-17 kcal/mol

Q. Heat of combustion $C (s), H_{2} (g)$ and $C H_{4} (g)$ are $- 94, - 68$ and $- 213 k c a l / m o l$ . Then $Δ H^{\circ}$ for $C (s) + 2 H_{2} (g) \to Δ C H_{4} (g)$ is

$- 17 k c a l$

$- 111 k c a l$

$- 170 k c a l$

$- 85 k c a l$

Q. Heat of combustion C(s),H2​(g) and CH4​(g) are −94,−68 and −213kcal/mol. Then ΔH∘ for C(s)+2H2​(g)→ΔCH4​(g)is

−17kcal

−111kcal

−170kcal

−85kcal

Q. Heat of combustion $C (s), H_{2} (g)$ and $C H_{4} (g)$ are $- 94, - 68$ and $- 213 k c a l / m o l$ . Then $Δ H^{\circ}$ for $C (s) + 2 H_{2} (g) \to Δ C H_{4} (g)$ is

$- 17 k c a l$

$- 111 k c a l$

$- 170 k c a l$

$- 85 k c a l$