(a) It cannot be made spontaneous without increasing the temperature.

(b) It can be made spontaneous by decreasing the pressure at 298 K.

(c) It can be made spontaneous by increasing the pressure at 298 K.

(d) None of the above

(a) It cannot be made spontaneous without increasing the temperature.

(b) It can be made spontaneous by decreasing the pressure at 298 K.

(c) It can be made spontaneous by increasing the pressure at 298 K.

(d) None of the above

Answer: It can be made spontaneous by decreasing the pressure at 298 K.

For the reaction to be spontaneous, $\Delta G = -ve$

At equilibrium, $\Delta G =0$

$\Delta G = \Delta G^\circ - 2.303$ RT log $(p_{O_2})^{\frac{1}{2}}$

$\Rightarrow 0 = +11210 - 2.303\times (8.31) \times 298$ log $(p_{O_2})^{\frac{1}{2}}$

Solving $(p_{O_2})$ = 0.000116 atm = 0.089 torr

Therefore, at a pressure of less than 0.089 torr, the reaction is spontaneous.

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