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Thermodynamics question

2005 · Shift 0 · Q2
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Thermodynamics question

2005 · Shift 0 · Q2

JEE MainChemistryThermodynamicsMCQ+4 / −1
Consider the reaction: N2N_2N2​ + 3H2H_2H2​ →\to→ 2NH3NH_3NH3​ carried out at constant temperature and pressure. If ΔH\Delta HΔH and ΔU\Delta UΔU are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?
  1. A
    ΔH\Delta HΔH>ΔU\Delta UΔU
  2. B
    ΔH\Delta HΔH<ΔU\Delta UΔU
  3. C
    ΔH\Delta HΔH=ΔU\Delta UΔU
  4. D
    ΔH\Delta HΔH = 0
View written solutionFree

Correct answer: B

  1. For gaseous reactions, the relation between enthalpy change and internal energy change is
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

where Δng\Delta n_gΔng​ is the change in moles of gaseous species:

Δng=(moles of gaseous products)−(moles of gaseous reactants)\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})Δng​=(moles of gaseous products)−(moles of gaseous reactants)
  1. For the reaction
N2+3H2→2NH3N_2 + 3H_2 \to 2NH_3N2​+3H2​→2NH3​

all species are gases, so

  • Reactant gas moles =1+3=4= 1 + 3 = 4=1+3=4
  • Product gas moles =2= 2=2

Hence,

Δng=2−4=−2\Delta n_g = 2 - 4 = -2Δng​=2−4=−2
  1. Substitute into the formula:
ΔH=ΔU+(−2)RT\Delta H = \Delta U + (-2)RTΔH=ΔU+(−2)RT ΔH=ΔU−2RT\Delta H = \Delta U - 2RTΔH=ΔU−2RT

Since RT>0RT > 0RT>0, it follows that

ΔH<ΔU\Delta H < \Delta UΔH<ΔU
  1. Therefore, the correct option is:
B: ΔH<ΔU\boxed{\text{B: } \Delta H < \Delta U}B: ΔH<ΔU​
  1. Comparison with stored correct answer:

Stored correct answer = B, which matches the derived answer.

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