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What will be the temperature at which liquid and vapour of water will be at equilibrium with each other? Given data are\(\triangle\)Hvap = 40.73 kJ mol-1\(\triangle\)Svap = 0.109 kJ mol-1 K-1(a) 273 K(b) 373 K(c) 37° C(d) 110°C |
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Answer» Answer is : (b) 373 K We know that at equilibrium, \(\triangle G=0\) for given reaction \(H_2O(l) \rightleftharpoons H_2O(v)\) At equilibrium, \(\triangle G=0\) \(\triangle G=\triangle H-T\triangle S=0\) \(\triangle H=T\triangle S\) \(T=\frac{\triangle H}{\triangle S}\) \(=\frac{40.73\, kJ\, mol^{-1}} {0.109 \,kJ\, mol^{-1}}\) = 373 K |
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