1.

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

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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