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Thermodynamics question

2022 · 28 Jun · Shift 2 · Q13
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Thermodynamics question

2022 · 28 Jun · Shift 2 · Q13

JEE MainChemistryThermodynamicsNumerical+4 / −1
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, Δ\DeltaΔ CH Θ\ThetaΘ=−-− 601.70 kJ mol −-− 1, the magnitude of change in internal energy for the reaction is ‾\underline{\hspace{2cm}}​ kJ. (Nearest integer) (Given : R = 8.3 J K −-− 1 mol −-− 1)
Numerical answer
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Correct answer: 600

  1. Write the combustion reaction

For combustion of 1 mole of magnesium:

Mg(s)+12O2(g)→MgO(s)\mathrm{Mg(s) + \frac{1}{2}O_2(g) \rightarrow MgO(s)}Mg(s)+21​O2​(g)→MgO(s)

  1. Relation between enthalpy change and internal energy change

For a reaction 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:

  • Reactant gas moles =12= \frac{1}{2}=21​
  • Product gas moles =0= 0=0

So,

Δng=0−12=−12\Delta n_g = 0 - \frac{1}{2} = -\frac{1}{2}Δng​=0−21​=−21​

  1. Substitute into the formula

Given:

  • ΔH=−601.70 kJ mol−1\Delta H = -601.70\ \text{kJ mol}^{-1}ΔH=−601.70 kJ mol−1
  • R=8.3 J K−1mol−1R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}R=8.3 J K−1mol−1
  • T=300 KT = 300\ \text{K}T=300 K

Using

ΔU=ΔH−ΔngRT\Delta U = \Delta H - \Delta n_g RTΔU=ΔH−Δng​RT

ΔU=−601.70−(−12)RT\Delta U = -601.70 - \left(-\frac{1}{2}\right)RTΔU=−601.70−(−21​)RT

ΔU=−601.70+12(8.3)(300)×10−3\Delta U = -601.70 + \frac{1}{2}(8.3)(300)\times 10^{-3}ΔU=−601.70+21​(8.3)(300)×10−3

(Converting J to kJ)

12(8.3)(300)=1245 J=1.245 kJ\frac{1}{2}(8.3)(300) = 1245\ \text{J} = 1.245\ \text{kJ}21​(8.3)(300)=1245 J=1.245 kJ

So,

ΔU=−601.70+1.245=−600.455 kJ\Delta U = -601.70 + 1.245 = -600.455\ \text{kJ}ΔU=−601.70+1.245=−600.455 kJ

  1. Magnitude of change in internal energy

∣ΔU∣=600.455 kJ|\Delta U| = 600.455\ \text{kJ}∣ΔU∣=600.455 kJ

Nearest integer:

600\boxed{600}600​

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