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Thermodynamics question

2018 · Shift 0 · Q23
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Thermodynamics question

2018 · Shift 0 · Q23

JEE MainChemistryThermodynamicsMCQ+4 / −1
The combustion of benzene(l) gives CO2CO_2CO2​(g) and H2OH_2OH2​O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be : (R = 8.314 JK–1 mol–1)
  1. A
    –3267.6
  2. B
    4152.6
  3. C
    –452.46
  4. D
    3260
View written solutionFree

Correct answer: A

  1. Write the combustion reaction

For benzene: C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)\mathrm{C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)}C6​H6​(l)+215​O2​(g)→6CO2​(g)+3H2​O(l)

  1. Relation between heat at constant pressure and constant volume

At constant volume: qv=ΔUq_v = \Delta Uqv​=ΔU

At constant pressure: qp=ΔHq_p = \Delta Hqp​=ΔH

For reactions 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:

  • Gaseous products = 666 mol of CO2CO_2CO2​
  • Gaseous reactants = 152=7.5\frac{15}{2} = 7.5215​=7.5 mol of O2O_2O2​

So, Δng=6−7.5=−1.5\Delta n_g = 6 - 7.5 = -1.5Δng​=6−7.5=−1.5

  1. Compute ΔngRT\Delta n_g RTΔng​RT

Given: R=8.314 J mol−1 K−1,T=25∘C=298 KR = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}, \quad T = 25^\circ C = 298\ \mathrm{K}R=8.314 Jmol−1K−1,T=25∘C=298 K

ΔngRT=(−1.5)(8.314)(298) J mol−1\Delta n_g RT = (-1.5)(8.314)(298)\ \mathrm{J\,mol^{-1}}Δng​RT=(−1.5)(8.314)(298) Jmol−1

=−3715.36 J mol−1= -3715.36\ \mathrm{J\,mol^{-1}}=−3715.36 Jmol−1

=−3.715 kJ mol−1= -3.715\ \mathrm{kJ\,mol^{-1}}=−3.715 kJmol−1

  1. Calculate heat of combustion at constant pressure

Given heat of combustion at constant volume: ΔU=−3263.9 kJ mol−1\Delta U = -3263.9\ \mathrm{kJ\,mol^{-1}}ΔU=−3263.9 kJmol−1

Thus, ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT ΔH=−3263.9+(−3.715)\Delta H = -3263.9 + (-3.715)ΔH=−3263.9+(−3.715) ΔH≈−3267.6 kJ mol−1\Delta H \approx -3267.6\ \mathrm{kJ\,mol^{-1}}ΔH≈−3267.6 kJmol−1

  1. Match with options

The correct option is: A: −3267.6 kJ mol−1\boxed{\text{A: } -3267.6\ \mathrm{kJ\,mol^{-1}}}A: −3267.6 kJmol−1​

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