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Thermodynamics question

2006 · Shift 0 · Q3
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Thermodynamics question

2006 · Shift 0 · Q3

JEE MainChemistryThermodynamicsMCQ+4 / −1
(ΔH−ΔU\Delta H - \Delta UΔH−ΔU) for the formation of carbon monoxide (CO) from its elements at 298 K is : (R = 8.314 J K–1 mol–1)
  1. A
    –1238.78 J mol–1
  2. B
    1238.78 J mol–1
  3. C
    –2477.57 J mol–1
  4. D
    2477.57 J mol–1
View written solutionFree

Correct answer: B

  1. Write the formation reaction of CO from its elements

For formation of 1 mole of carbon monoxide:

C(s)+12O2(g)→CO(g)\text{C(s)} + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CO(g)}C(s)+21​O2​(g)→CO(g)

  1. Use the relation between enthalpy and internal energy

For a reaction involving gases:

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.

  1. Calculate Δng\Delta n_gΔng​

From the reaction:

  • Reactant gaseous moles =12= \frac{1}{2}=21​
  • Product gaseous moles =1= 1=1

So,

Δng=1−12=12\Delta n_g = 1 - \frac{1}{2} = \frac{1}{2}Δng​=1−21​=21​

  1. Substitute the values

Given:

  • R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}R=8.314 J K−1mol−1
  • T=298 KT = 298\ \text{K}T=298 K

Hence,

ΔH−ΔU=12×8.314×298\Delta H - \Delta U = \frac{1}{2} \times 8.314 \times 298ΔH−ΔU=21​×8.314×298

=0.5×2477.572= 0.5 \times 2477.572=0.5×2477.572

=1238.786 J mol−1= 1238.786\ \text{J mol}^{-1}=1238.786 J mol−1

  1. Match with the options

ΔH−ΔU≈1238.78 J mol−1\Delta H - \Delta U \approx 1238.78\ \text{J mol}^{-1}ΔH−ΔU≈1238.78 J mol−1

So the correct option is B.

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