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Chemistry Question on Thermodynamics

Heat of combustion ΔH∘ \Delta H^{\circ} for C(s),H2(g) C ( s ), H _{2}( g ) and CH4(g)CH _{4}( g ) are -94,-68 and −213 Kcal/mol-213 \,Kcal / mol then ΔH∘\Delta H ^{\circ} for C(s)+2H2(g)→CH4(g)C ( s )+2 H _{2}( g ) \to CH _{4}( g ) is:-

A

- 17 Kcal

B

- 111 Kcal

C

- 170 Kcal

D

- 85 Kcal

Answer

- 17 Kcal

Explanation

Solution

For reaction ,
C(s)+2H2(g)→CH4(g),ΔH∘=?C (s) + 2H_2 (g) \to CH_4(g), \Delta H^\circ = ?
ΔH∘=−[(H∘of  combustion  ofC+2×ΔHof  combustion  ofH2)]\Delta H^\circ = - [(H^\circ of \,\, combustion\,\,of C + 2\times \Delta H of\,\,combustion \,\, of H_2)]
C+O2→CC2;ΔH=−94  kcalC + O_2 \to CC_2; \Delta H = - 94\,\, kcal
2H2+O2→2H2O;ΔH=−68×2  kcal?2H_2 +O_2 \to 2H_2O; \Delta H = - 68 \times 2\,\, kcal?
CH4+2O2→CO2+2H2O;Δ=−213  kcalCH_4 +2O_2 \to CO_2+ 2H_2O; \Delta = - 213 \,\,kcal
=−[(−213)−(−94+2×−68)]Kcal/mol= - [(-213) - (-94+2 \times - 68)] Kcal/mol
=−[−213+230]=−17kcal/mol= - [-213+230] = - 17 kcal/mol