Question

# The enthalpy of vapourization of a liquid is $$30 \,kJ \,mol^{-1}$$ and entropy of vapourization is $$75 \,J \,mol^{-1}K$$. The boiling point of the liquid at $$1 \,atm$$ is:

A
250K
B
400K
C
450K
D
600K

Solution

## The correct option is A $$400 \,K$$For the equilibrium process:$$T = \dfrac{\Delta H_{vap}}{\Delta S}$$$$= \dfrac{30 \times 10^3 \,J \,mol^{-1}}{75 \,J \,mol^{-1}}$$$$T = 400 \,K$$Chemistry

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