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For a reaction R(1), Delta G = x KJ mol^...

For a reaction `R_(1), Delta G = x KJ mol^(-1)`. For a reaction `R_(2), Delta G = y KJ mol^(-1)`. Reaction `R_(1)` is non-spontaneous but along with `R_(2)` it is spontaneous. This means that

A

x is -ve, y is +ve but in magnitude `x gt y`

B

x is +ve, y is -ve but in magnitude `y gt x`

C

Both x and y are -ve but not equal

D

Both x and y +ve but not equal

Text Solution

Verified by Experts

The correct Answer is:
B
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Knowledge Check

  • For A rarr B, DeltaH= 3.5 kcal mol^(-1), DeltaS= 10 cal mol^(-1) K^(-1) . Reaction is non spontaneous at

    A
    400K
    B
    `27^(@)C`
    C
    `77^(@)C`
    D
    350K

  • Ractants rarr products Delta G = + 63 KJ "mole"^(-1) This reaction is made to take plane by coupling with

    A
    `ATP overset(H_2O) ADP`
    B
    `ATP overset(H_2O) rarr A`
    C
    `ATP overset(H_2O) rarr AMP`
    D
    `ADP overset(H_2O) rarr A`

  • At the temperature T(K) for the reaction X_(2)O_(4) (l) rarr 2XO_(2)(g) Delta U = x kJ mol^(-1), Delta S = y JK^(-1) mol^(-1) . Gibbs energy change for the reaction is (Assume X_(2)O_(4), XO_(2) are ideal gases) Delta U = x kJ mol^(-1), Delta S = y JK^(-1) mol^(-1)

    A
    `1000 x + 2R(T - y) Jmol^(-1)`
    B
    `1000 x + T(2R-y) J mol^(-1)`
    C
    `x + T (2R - y) J mol^(-1)`
    D
    `x + 2R(T - y) J mol^(-1)`