1) Calculate the change in the entropy of the system, the surroundings, and the

Question

1) Calculate the change in the entropy of the system, the surroundings, and the universe, when 1 L of an ideal gas at 298 K and 2 bar is expanded to 2 L at the same temperature via (a) an isothermal reversible expansion (while matching the pressure on the gas to that of the ideal gas), (b) an irreversible expansion against p 1 bar Calculate the change in entropy of the system, the surroundings and the universe for the following reaction under standard conditions at 298.15 K Na (s) + H2O (l)-> NaOH (aq) + ½ H2 (g) Consider reactants and products to be the system, and use the following data Na (s) H20 (I NaOH (aq Hz (g) Hr at 298 K kJ mol1 0 285.83 -469.15 0 S° at 298 K J K1 mol 51.21 69.95 48.1 130.7 State whether the reaction is entropically or enthalpically favoured, or both

Explanation / Answer

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(a) Isothermal expansion,

q = - w = n R T ln(V2/V1) = - 1 * 298 * 8.314 * ln (2/1) = - 1717 J

Entropy change = q / T = 1717 / 298 = 5.76 J.K-1

deltaH0r = deltaH0f(NaOH) + (1/2) deltaH0f(H2) - deltaH0f(Na) - deltaH0f(H2O)

= - 469.15 + (1/2)(0) - (0) - ( - 285.83)

= - 183.32 kJ

deltaS0r = deltaS0f(NaOH) + (1/2) deltaS0f(H2) - deltaS0f(Na) - deltaS0f(H2O)

= 48.1 + (1/2)(130.7) - 51.21 - 69.95

= - 7.71 J

Since, deltaH0f is negative, it is enthalpically favourable.

But, deltaS0r is negative, so it not entropically favourable.

(For entropically favourable reaction deltaS0r should be positive)

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