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NEET CHEMISTRYMedium

Heat of combustion ΔH∘\Delta H^\circ for C(s)C(s), H2(g)H_2(g) and CH4(g)CH_4(g) are −94-94, −68-68 and −213 Kcal/mol-213\text{ Kcal/mol}. ΔH∘\Delta H^\circ for C(s)+2H2(g)→CH4(g)C(s) + 2H_2(g) \rightarrow CH_4 (g) is:

A

−17 Kcal-17\text{ Kcal}

B

−111 Kcal-111\text{ Kcal}

C

−170 Kcal-170\text{ Kcal}

D

−85 Kcal-85\text{ Kcal}

Step-by-Step Solution

  1. Identify the Goal: Calculate the enthalpy change (ΔHreaction\Delta H_{reaction}) for the formation of methane (CH4CH_4) using heat of combustion data. Reaction: C(s)+2H2(g)→CH4(g)C(s) + 2H_2(g) \rightarrow CH_4(g)
  2. Apply Formula: When heat of combustion (ΔHc\Delta H_c) data is given, the enthalpy of reaction is calculated as: ΔHreaction=∑ΔHc(Reactants)−∑ΔHc(Products)\Delta H_{reaction} = \sum \Delta H_c(\text{Reactants}) - \sum \Delta H_c(\text{Products})
  3. Substitute Values:
  • Reactants: 1 mole of C(s)C(s) and 2 moles of H2(g)H_2(g).
  • Product: 1 mole of CH4(g)CH_4(g). ΔHreaction=[1⋅ΔHc(C)+2⋅ΔHc(H2)]−[1⋅ΔHc(CH4)]\Delta H_{reaction} = [1 \cdot \Delta H_c(C) + 2 \cdot \Delta H_c(H_2)] - [1 \cdot \Delta H_c(CH_4)] ΔHreaction=[(−94)+2(−68)]−[−213]\Delta H_{reaction} = [(-94) + 2(-68)] - [-213]
  1. Calculate: ΔHreaction=[−94−136]+213\Delta H_{reaction} = [-94 - 136] + 213 ΔHreaction=−230+213=−17 Kcal\Delta H_{reaction} = -230 + 213 = -17\text{ Kcal}
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