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Thermodynamics question

2007 · Shift 0 · Q8
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Thermodynamics question

2007 · Shift 0 · Q8

JEE MainChemistryThermodynamicsMCQ+4 / −1
In conversion of lime-stone to lime, CaCO3CaCO_3CaCO3​(s) →\to→ CaOCaOCaO(s) + CO2CO_2CO2​ (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :
  1. A
    1008 K
  2. B
    1200
  3. C
    845 K
  4. D
    1118 K
View written solutionFree

Correct answer: D

  1. Condition for spontaneity

For the reaction CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \to CaO(s) + CO_2(g)CaCO3​(s)→CaO(s)+CO2​(g) spontaneity is determined by Gibbs free energy: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

The reaction becomes spontaneous when ΔG<0\Delta G < 0ΔG<0 The threshold temperature is obtained from ΔG=0\Delta G = 0ΔG=0 so, T=ΔHΔST = \frac{\Delta H}{\Delta S}T=ΔSΔH​

  1. Given data

ΔH∘=+179.1 kJ mol−1\Delta H^\circ = +179.1\, \text{kJ mol}^{-1}ΔH∘=+179.1kJ mol−1 ΔS∘=160.2 J K−1mol−1\Delta S^\circ = 160.2\, \text{J K}^{-1}\text{mol}^{-1}ΔS∘=160.2J K−1mol−1

Convert enthalpy into joules: 179.1 kJ mol−1=179100 J mol−1179.1\, \text{kJ mol}^{-1} = 179100\, \text{J mol}^{-1}179.1kJ mol−1=179100J mol−1

  1. Calculate the equilibrium temperature

T=179100160.2T = \frac{179100}{160.2}T=160.2179100​

Now compute: T≈1117.98 KT \approx 1117.98\, \text{K}T≈1117.98K

So, T≈1118 KT \approx 1118\, \text{K}T≈1118K

  1. Interpretation
  • At T=1118 KT = 1118\, \text{K}T=1118K, ΔG=0\Delta G = 0ΔG=0
  • For temperatures above this value, ΔG<0\Delta G < 0ΔG<0 and the reaction is spontaneous.
  1. Option check
  • A: 1008 K1008\,\text{K}1008K — incorrect
  • B: 1200 K1200\,\text{K}1200K — this is above threshold, but the question asks for the temperature above which spontaneity starts, i.e. the threshold value
  • C: 845 K845\,\text{K}845K — incorrect
  • D: 1118 K1118\,\text{K}1118K — correct

Therefore, the correct answer is D.

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