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Thermodynamics question

2019 · 12 Jan · Shift 2 · Q14
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Thermodynamics question

2019 · 12 Jan · Shift 2 · Q14

JEE MainChemistryThermodynamicsMCQ+4 / −1
Given (i) C (graphite) + O2O_2O2​(g) →\to→ CO2CO_2CO2​(g); Δ\DeltaΔ rH Θ^\ThetaΘ = x kJ mol −-− 1 (ii) C(graphite) + 12{1 \over 2}21​ O2O_2O2​(g) →\to→ COCOCO(g); Δ\DeltaΔ rH Θ^\ThetaΘ = y kJ mol −-− 1 (iii) COCOCO(g) + 12{1 \over 2}21​ O2O_2O2​(g) →\to→ CO2CO_2CO2​(g); Δ\DeltaΔ rH Θ^\ThetaΘ = z kJ mol −-− 1 Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?
  1. A
    z = x + y
  2. B
    x = y + z
  3. C
    x = y – z
  4. D
    y = 2z – x
View written solutionFree

Correct answer: B

  1. Write the given reactions

    (i)C(graphite)+O2(g)→CO2(g)ΔrHΘ=x\text{(i)}\quad \mathrm{C(graphite)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)} \qquad \Delta_r H^\Theta = x(i)C(graphite)+O2​(g)→CO2​(g)Δr​HΘ=x

    (ii)C(graphite)+12O2(g)→CO(g)ΔrHΘ=y\text{(ii)}\quad \mathrm{C(graphite)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CO(g)} \qquad \Delta_r H^\Theta = y(ii)C(graphite)+21​O2​(g)→CO(g)Δr​HΘ=y

    (iii)CO(g)+12O2(g)→CO2(g)ΔrHΘ=z\text{(iii)}\quad \mathrm{CO(g)} + \frac{1}{2}\mathrm{O_2(g)} \to \mathrm{CO_2(g)} \qquad \Delta_r H^\Theta = z(iii)CO(g)+21​O2​(g)→CO2​(g)Δr​HΘ=z

  2. Apply Hess's law

    If we add reactions (ii) and (iii), we get:

    (C+12O2→CO)+(CO+12O2→CO2)\left(\mathrm{C} + \frac{1}{2}\mathrm{O_2} \to \mathrm{CO}\right) + \left(\mathrm{CO} + \frac{1}{2}\mathrm{O_2} \to \mathrm{CO_2}\right)(C+21​O2​→CO)+(CO+21​O2​→CO2​)

    Cancelling CO\mathrm{CO}CO on both sides:

    C(graphite)+O2(g)→CO2(g)\mathrm{C(graphite)} + \mathrm{O_2(g)} \to \mathrm{CO_2(g)}C(graphite)+O2​(g)→CO2​(g)

    This is exactly reaction (i).

  3. Relate the enthalpy changes

    Therefore,

    x=y+zx = y + zx=y+z

  4. Match with the options

    The correct option is:

    B: x=y+z\boxed{\text{B: } x = y + z}B: x=y+z​

  5. Compare with stored answer

    Stored correct answer: B

    Our derived answer: B

    So they agree.

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