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Thermodynamics question

2020 · Q81
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Thermodynamics question

2020 · Q81

NEETChemistryThermodynamicsMCQ+4 / −1
Hydrolysis of sucrose is given by the following reaction.
Sucrose + H2O ⇌ Glucose + Fructose
If the equilibrium constant (Kc) is 2 ×\times× 1013 at 800 K, the value of ΔrGΘ\Delta r{G^\Theta }ΔrGΘ at the same temperature will be :
  1. A
    8.314 J mol−1K−1×300K×In(2×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})mol−1K−1×300K×In(2×1013)
  2. B
    8.314 J mol−1K−1×300K×In(3×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(3 \times {10^{13}})mol−1K−1×300K×In(3×1013)
  3. C
    −-− 8.314 J mol−1K−1×300K×In(4×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(4 \times {10^{13}})mol−1K−1×300K×In(4×1013)
  4. D
    -8.314 J mol−1K−1×300K×In(2×1013)mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})mol−1K−1×300K×In(2×1013)
View written solutionFree

Correct answer: D

ΔG\Delta GΔG = ΔG\Delta GΔG ∘^\circ ∘

  • RT ln Q

    At equilibrium ΔG\Delta GΔG = 0, Q = Keq

    So, ΔrG∘{\Delta _r}G^\circ Δr​G∘ = −-− RT in Keq

    ΔrG∘{\Delta _r}G^\circ Δr​G∘ = −-− 8.314 J mol-1 K-1 ×\times× 300K ×\times× ln (2 ×\times× 1013)
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