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Thermodynamics question

2019 · 10 Apr · Shift 2 · Q7
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Thermodynamics question

2019 · 10 Apr · Shift 2 · Q7

JEE MainChemistryThermodynamicsMCQ+4 / −1
The difference between Δ\DeltaΔ H and Δ\DeltaΔ U (Δ\DeltaΔ H – Δ\DeltaΔ U), when the combustion of one mole of heptane(l) is carried out at a temperature T, is equal to :
  1. A
    – 4 RT
  2. B
    3 RT
  3. C
    – 3 RT
  4. D
    4 RT
View written solutionFree

Correct answer: A

  1. Use the relation between enthalpy and internal energy

For a reaction involving gases,

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

So,

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:

Δng=(moles of gaseous products)−(moles of gaseous reactants)\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})Δng​=(moles of gaseous products)−(moles of gaseous reactants)
  1. Write the combustion reaction of one mole of heptane

Heptane is C7H16(l)\mathrm{C_7H_{16}(l)}C7​H16​(l).

Its combustion is:

C7H16(l)+11 O2(g)→7 CO2(g)+8 H2O(l)\mathrm{C_7H_{16}(l) + 11\,O_2(g) \rightarrow 7\,CO_2(g) + 8\,H_2O(l)}C7​H16​(l)+11O2​(g)→7CO2​(g)+8H2​O(l)

At temperature TTT, unless otherwise stated, water formed in combustion is taken as liquid because the reactant is liquid heptane and standard combustion products are CO2(g)\mathrm{CO_2(g)}CO2​(g) and H2O(l)\mathrm{H_2O(l)}H2​O(l).

  1. Calculate Δng\Delta n_gΔng​

Gaseous reactants:

11 mol O211 \text{ mol } O_211 mol O2​

Gaseous products:

7 mol CO27 \text{ mol } CO_27 mol CO2​

Therefore,

Δng=7−11=−4\Delta n_g = 7 - 11 = -4Δng​=7−11=−4
  1. Find ΔH−ΔU\Delta H - \Delta UΔH−ΔU
ΔH−ΔU=ΔngRT=(−4)RT\Delta H - \Delta U = \Delta n_g RT = (-4)RTΔH−ΔU=Δng​RT=(−4)RT

So,

ΔH−ΔU=−4RT\boxed{\Delta H - \Delta U = -4RT}ΔH−ΔU=−4RT​
  1. Check options
  • A: −4RT-4RT−4RT ✅
  • B: 3RT3RT3RT ❌
  • C: −3RT-3RT−3RT ❌
  • D: 4RT4RT4RT ❌

Hence, the correct option is A.

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