Question

If enthalpy of vaporisation of water is $186.5kJ/mol$, the entropy of its vaporisation will be:

• A. $0.5$
• B. $1.0$
• C. $1.5$
• D. $2.0$

Answer 1

Answer: (A)

$0.5$

Since, for any reaction, the entropy is given by,

$Entropy=\frac{\Delta H_{process}}{Temperature\left(\text{in Kelvin}\right)}$

Boiling point of water = ${100}^{0}C$ $+273K$=$373K$

hence, $entropy=\frac{186.5}{373}=0.5$ (Since, vaporisation occurs at 100*C = 373K)

Hence, answer is option $A$.