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Thermodynamics question

2012 · Shift 2 · Q10
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Thermodynamics question

2012 · Shift 2 · Q10

JEE AdvancedChemistryThermodynamicsMCQ+3 / −1
Using the data provided, calculate the multiple bond energy (kJ mol -1) of a C≡C bond in C2H2C_2H_2C2​H2​. That energy is (take the bond energy of C-H bond as 350 kJ mol -1). 2C(s)+H2(g)→C2H2ΔH=225 kJ mol−12C(s)→2C(g)ΔH=1410 kJ mol−1H2(g)→2H(g)ΔH=330 kJ mol−1\begin{array}{ll} 2\text{C}(s) + \text{H}_2(g) \to \text{C}_2\text{H}_2 & \Delta H = 225 \text{ kJ mol}^{-1} \\ 2\text{C}(s) \to 2\text{C}(g) & \Delta H = 1410 \text{ kJ mol}^{-1} \\ \text{H}_2(g) \to 2\text{H}(g) & \Delta H = 330 \text{ kJ mol}^{-1} \end{array}2C(s)+H2​(g)→C2​H2​2C(s)→2C(g)H2​(g)→2H(g)​ΔH=225 kJ mol−1ΔH=1410 kJ mol−1ΔH=330 kJ mol−1​
  1. A
    1165 kJ mol −-− 1
  2. B
    837 kJ mol −-− 1
  3. C
    865 kJ mol −-− 1
  4. D
    815 kJ mol −-− 1
View written solutionFree

Correct answer: D

  1. Use Hess's law to find the atomization enthalpy of C2H2\mathrm{C_2H_2}C2​H2​.

Given: 2C(s)+H2(g)→C2H2(g),ΔH=+225 kJ mol−12\mathrm{C}(s)+\mathrm{H_2}(g)\to \mathrm{C_2H_2}(g),\quad \Delta H=+225\ \text{kJ mol}^{-1}2C(s)+H2​(g)→C2​H2​(g),ΔH=+225 kJ mol−1

Also, 2C(s)→2C(g),ΔH=1410 kJ mol−12\mathrm{C}(s)\to 2\mathrm{C}(g),\quad \Delta H=1410\ \text{kJ mol}^{-1}2C(s)→2C(g),ΔH=1410 kJ mol−1 H2(g)→2H(g),ΔH=330 kJ mol−1\mathrm{H_2}(g)\to 2\mathrm{H}(g),\quad \Delta H=330\ \text{kJ mol}^{-1}H2​(g)→2H(g),ΔH=330 kJ mol−1

So, converting the elements in their standard states to gaseous atoms: 2C(s)+H2(g)→2C(g)+2H(g)2\mathrm{C}(s)+\mathrm{H_2}(g)\to 2\mathrm{C}(g)+2\mathrm{H}(g)2C(s)+H2​(g)→2C(g)+2H(g) ΔH=1410+330=1740 kJ mol−1\Delta H = 1410+330 = 1740\ \text{kJ mol}^{-1}ΔH=1410+330=1740 kJ mol−1

  1. Now relate this to bond formation in acetylene.

From gaseous atoms to acetylene: 2C(g)+2H(g)→HC≡CH(g)2\mathrm{C}(g)+2\mathrm{H}(g)\to \mathrm{HC\equiv CH}(g)2C(g)+2H(g)→HC≡CH(g) Let the bond energy of C≡C\mathrm{C\equiv C}C≡C be xxx.

In C2H2\mathrm{C_2H_2}C2​H2​, bonds present are:

  • one C≡C\mathrm{C\equiv C}C≡C bond: energy xxx
  • two C−H\mathrm{C-H}C−H bonds: energy 2×350=7002\times 350 = 7002×350=700

Hence, bond formation releases: x+700x+700x+700 So the enthalpy change for atom combination is: ΔH=−(x+700)\Delta H = -(x+700)ΔH=−(x+700)

  1. Apply Hess's law for the overall formation reaction.

Overall: 2C(s)+H2(g)→C2H2(g)2\mathrm{C}(s)+\mathrm{H_2}(g)\to \mathrm{C_2H_2}(g)2C(s)+H2​(g)→C2​H2​(g)

This can be written as:

  • atomization of reactants: +1740+1740+1740
  • formation of bonds in product: −(x+700)-(x+700)−(x+700)

Therefore, 225=1740−(x+700)225 = 1740 - (x+700)225=1740−(x+700)

  1. Solve for xxx.

225=1740−x−700225 = 1740 - x - 700225=1740−x−700 225=1040−x225 = 1040 - x225=1040−x x=1040−225=815 kJ mol−1x = 1040 - 225 = 815\ \text{kJ mol}^{-1}x=1040−225=815 kJ mol−1

  1. Match with the options.

815 kJ mol−1\boxed{815\ \text{kJ mol}^{-1}}815 kJ mol−1​ So, the correct option is D.

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