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Determine whether the following reaction will be spontaneous or non-spontaneous under standard conditions.Zn(s) + Cu2+ → Zn2+ +Cu(s) ΔH0 = -219 kJ, ΔS0 = -21 JK-1 |
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Answer» Given : ΔH0 = -219 kJ; ΔS0 = -21 JK-1 = 0.021 kJ K-1 ΔG0 = ? For standard conditions : Pressure = 1 atm Temperature = T = 298 K ΔG0 = ΔH0 - TΔS0 = -219 – 298 × (-0.021) = -219 + 6.258 = -212.742 kJ Since ΔG < 0, ∴ The reaction is spontaneous. |
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