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NEET CHEMISTRYThermodynamicsMedium

Question

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=[1ΔHc(C)+2ΔHc(H2)][1ΔHc(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=[94136]+213\Delta H_{reaction} = [-94 - 136] + 213 ΔHreaction=230+213=17 Kcal\Delta H_{reaction} = -230 + 213 = -17\text{ Kcal}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicscombustionkcalmolrightarrow

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