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Thermodynamics question

2025 · 3 Apr · Shift 1 · Q25
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Thermodynamics question

2025 · 3 Apr · Shift 1 · Q25

JEE MainChemistryThermodynamicsNumerical+4 / −1
Given : ΔH⊖sub [C (graphite )]=710 kJ mol−1ΔC−HH⊖=414 kJ mol−1ΔH−HH⊖=436 kJ mol−1ΔC=CH⊖=611 kJ mol−1\begin{aligned} & \left.\Delta \mathrm{H}^{\ominus}{ }_{\text {sub }}[\mathrm{C} \text { (graphite })\right]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}}=\mathrm{C} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}​ΔH⊖sub ​[C (graphite )]=710 kJ mol−1ΔC−H​H⊖=414 kJ mol−1ΔH−H​H⊖=436 kJ mol−1ΔC​=CH⊖=611 kJ mol−1​ The ΔHf⊖\Delta \mathrm{H}_{\mathrm{f}} \ominusΔHf​⊖ for CH2=CH2\mathrm{CH}_2=\mathrm{CH}_2CH2​=CH2​ is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1 (nearest integer value)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Target reaction for standard enthalpy of formation

We need

2 C(graphite)+2 H2(g)→CH2=CH2(g)2\,\mathrm{C(graphite)} + 2\,\mathrm{H_2(g)} \rightarrow \mathrm{CH_2{=}CH_2(g)}2C(graphite)+2H2​(g)→CH2​=CH2​(g)

and must find

ΔHf∘[CH2=CH2].\Delta H_f^\circ[\mathrm{CH_2{=}CH_2}].ΔHf∘​[CH2​=CH2​].
  1. Use Hess's law via atomization

First, convert the elements into gaseous atoms.

  • For carbon:
2 C(graphite)→2 C(g)2\,\mathrm{C(graphite)} \rightarrow 2\,\mathrm{C(g)}2C(graphite)→2C(g) ΔH=2×710=1420 kJ mol−1\Delta H = 2\times 710 = 1420\ \mathrm{kJ\,mol^{-1}}ΔH=2×710=1420 kJmol−1
  • For hydrogen:
2 H2(g)→4 H(g)2\,\mathrm{H_2(g)} \rightarrow 4\,\mathrm{H(g)}2H2​(g)→4H(g)

Since

ΔHH−H∘=436 kJ mol−1,\Delta H_{\mathrm{H-H}}^\circ = 436\ \mathrm{kJ\,mol^{-1}},ΔHH−H∘​=436 kJmol−1,

for 2 moles of H2\mathrm{H_2}H2​,

ΔH=2×436=872 kJ mol−1\Delta H = 2\times 436 = 872\ \mathrm{kJ\,mol^{-1}}ΔH=2×436=872 kJmol−1

So total atomization enthalpy of reactants is

1420+872=2292 kJ mol−1.1420 + 872 = 2292\ \mathrm{kJ\,mol^{-1}}.1420+872=2292 kJmol−1.
  1. Form ethene from gaseous atoms

Ethene, CH2=CH2\mathrm{CH_2{=}CH_2}CH2​=CH2​, contains:

  • 444 C–H bonds
  • 111 C=C bond

Bond formation releases energy equal to bond enthalpy.

Hence energy released:

4(414)+611=1656+611=2267 kJ mol−1.4(414) + 611 = 1656 + 611 = 2267\ \mathrm{kJ\,mol^{-1}}.4(414)+611=1656+611=2267 kJmol−1.

Therefore,

2 C(g)+4 H(g)→CH2=CH2(g)2\,\mathrm{C(g)} + 4\,\mathrm{H(g)} \rightarrow \mathrm{CH_2{=}CH_2(g)}2C(g)+4H(g)→CH2​=CH2​(g)

has

ΔH=−2267 kJ mol−1.\Delta H = -2267\ \mathrm{kJ\,mol^{-1}}.ΔH=−2267 kJmol−1.
  1. Add the steps

Thus,

ΔHf∘=2292−2267=25 kJ mol−1.\Delta H_f^\circ = 2292 - 2267 = 25\ \mathrm{kJ\,mol^{-1}}.ΔHf∘​=2292−2267=25 kJmol−1.
  1. Final answer
25 kJ mol−1\boxed{25\ \mathrm{kJ\,mol^{-1}}}25 kJmol−1​

This matches the stored correct answer.

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