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Thermodynamics question

2005 · Shift 0 · Q3
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Thermodynamics question

2005 · Shift 0 · Q3

JEE MainChemistryThermodynamicsMCQ+4 / −1
If the bond dissociation energies of XYXYXY, X2X_2X2​ and Y2Y_2Y2​ (all diatomic molecules) are in the ratio of 1:1:0.5 and ΔHf\Delta H_fΔHf​ for the formation of XYXYXY is -200 kJ mole-1. The bond dissociation energy of X2X_2X2​ will be :
  1. A
    100 kJ mol-1
  2. B
    200 kJ mol-1
  3. C
    300 kJ mol-1
  4. D
    800 kJ mol-1
View written solutionFree

Correct answer: D

  1. Let the bond dissociation energies be: D(XY)=k, D(X2)=k, D(Y2)=0.5kD(XY)=k,\, D(X_2)=k,\, D(Y_2)=0.5kD(XY)=k,D(X2​)=k,D(Y2​)=0.5k since the ratio is given as XY:X2:Y2=1:1:0.5XY : X_2 : Y_2 = 1:1:0.5XY:X2​:Y2​=1:1:0.5

  2. The standard enthalpy of formation of XYXYXY is for the reaction: 12X2+12Y2→XY\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightarrow XY21​X2​+21​Y2​→XY with ΔHf=−200 kJ mol−1\Delta H_f = -200\ \text{kJ mol}^{-1}ΔHf​=−200 kJ mol−1

  3. Using bond energies, ΔH=Bonds broken−Bonds formed\Delta H = \text{Bonds broken} - \text{Bonds formed}ΔH=Bonds broken−Bonds formed

    Bonds broken: 12D(X2)+12D(Y2)=12k+12(0.5k)=0.5k+0.25k=0.75k\frac{1}{2}D(X_2) + \frac{1}{2}D(Y_2) = \frac{1}{2}k + \frac{1}{2}(0.5k)=0.5k+0.25k=0.75k21​D(X2​)+21​D(Y2​)=21​k+21​(0.5k)=0.5k+0.25k=0.75k

    Bonds formed: D(XY)=kD(XY)=kD(XY)=k

  4. Therefore, ΔHf=0.75k−k=−0.25k\Delta H_f = 0.75k - k = -0.25kΔHf​=0.75k−k=−0.25k

    Given: −0.25k=−200-0.25k = -200−0.25k=−200

    So, k=800 kJ mol−1k = 800\ \text{kJ mol}^{-1}k=800 kJ mol−1

  5. Since D(X2)=kD(X_2)=kD(X2​)=k we get D(X2)=800 kJ mol−1D(X_2)=800\ \text{kJ mol}^{-1}D(X2​)=800 kJ mol−1

  6. Checking options:

    • A: 100100100 ❌
    • B: 200200200 ❌
    • C: 300300300 ❌
    • D: 800800800 ✅

Hence, the correct answer is D.

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