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Thermodynamics question

2018 · 15 Apr · Shift 1 · Q20
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Thermodynamics question

2018 · 15 Apr · Shift 1 · Q20

JEE MainChemistryThermodynamicsMCQ+4 / −1
For which of the following reactions, Δ\DeltaΔ H is equal to Δ\DeltaΔ U?
  1. A
    N2N_2N2​(g) + 3H2H_2H2​(g) →\to→ 2NH3NH_3NH3​(g)
  2. B
    2HIHIHI(g) →\to→ H2H_2H2​(g) + I2I_2I2​(g)
  3. C
    2NO2NO_2NO2​(g) →\to→ N2O4N_2O_4N2​O4​(g)
  4. D
    2SO2SO_2SO2​(g) + O2O_2O2​(g) →\to→ 2SO3SO_3SO3​(g)
View written solutionFree

Correct answer: B

  1. For reactions involving gases, 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 ΔH=ΔU\Delta H = \Delta UΔH=ΔU, we must have:
Δng=0\Delta n_g = 0Δng​=0

Now check each option.

  1. Option A:
N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \to 2NH_3(g)N2​(g)+3H2​(g)→2NH3​(g)

Reactant gas moles =1+3=4= 1+3=4=1+3=4

Product gas moles =2=2=2

Δng=2−4=−2≠0\Delta n_g = 2-4 = -2 \neq 0Δng​=2−4=−2=0

So, ΔH≠ΔU\Delta H \ne \Delta UΔH=ΔU.

  1. Option B:
2HI(g)→H2(g)+I2(g)2HI(g) \to H_2(g) + I_2(g)2HI(g)→H2​(g)+I2​(g)

Reactant gas moles =2=2=2

Product gas moles =1+1=2=1+1=2=1+1=2

Δng=2−2=0\Delta n_g = 2-2=0Δng​=2−2=0

So,

ΔH=ΔU\Delta H = \Delta UΔH=ΔU
  1. Option C:
2NO2(g)→N2O4(g)2NO_2(g) \to N_2O_4(g)2NO2​(g)→N2​O4​(g)

Reactant gas moles =2=2=2

Product gas moles =1=1=1

Δng=1−2=−1≠0\Delta n_g = 1-2=-1 \neq 0Δng​=1−2=−1=0

So, ΔH≠ΔU\Delta H \ne \Delta UΔH=ΔU.

  1. Option D:
2SO2(g)+O2(g)→2SO3(g)2SO_2(g) + O_2(g) \to 2SO_3(g)2SO2​(g)+O2​(g)→2SO3​(g)

Reactant gas moles =2+1=3=2+1=3=2+1=3

Product gas moles =2=2=2

Δng=2−3=−1≠0\Delta n_g = 2-3=-1 \neq 0Δng​=2−3=−1=0

So, ΔH≠ΔU\Delta H \ne \Delta UΔH=ΔU.

  1. Therefore, the only reaction for which ΔH=ΔU\Delta H = \Delta UΔH=ΔU is:
B\boxed{\text{B}}B​
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