Question

For vaporization of water a \[1\] atmospheric pressure, the values of \[\Delta H\] and \[\Delta S\] are \[40.63kJmo{{l}^{-1}}\] and \[108.8J{{K}^{-1}}mo{{l}^{-1}}\], respectively. The temperature when Gibb’s energy change \[(\Delta G)\] for this transformation will be zero, is:
A \[273.4K\]
B \[393.4K\]
C \[373.4K\]
D \[293.4K\]

Answer 1

Hint: Gibbs free energy, also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Gibbs free energy is a state function hence it doesn’t depend on the path. So change in Gibbs free energy is equal to the change in enthalpy minus the product of temperature and entropy change of the system. The equation is given below
\[\Delta G=\Delta H-T\Delta S\],
\[\Delta G=\]Change in Gibbs’ free energy
\[\Delta H=\]Change in enthalpy
\[T=\]Change in temperature
\[ \Delta S=\]Change in entropy

Complete step by step answer:
Here, we have given that the values of change in enthalpy and entropy are \[40.63kJmo{{l}^{-1}}\] and \[108.8J{{K}^{-1}}mo{{l}^{-1}}\] respectively. Also, given the change in Gibbs energy is equal to zero and we have to find the corresponding temperature for this reaction (vaporization of water).
So, we have to substitute these values in the equation, i.e. \[\Rightarrow \Delta G=\Delta H-T\Delta S\]
Hence, we get;
\[ 0=\Delta H-T\Delta S\], so we can write it as
\[ \Delta H=T\Delta S\]
\[ T=\dfrac{\Delta H}{\Delta S}\]
\[\Rightarrow T=\frac{40.63\times {{10}^{-3}}Jmo{{l}^{-1}}}{108.8J{{K}^{-1}}mo{{l}^{-1}}}
=373.43K\]

So, the correct answer is Option C.

Note: Here the change in enthalpy is given in the unit of \[kJmo{{l}^{-1}}\].So, we have to convert this value into \[Jmo{{l}^{-1}}\] by multiplying the value with \[{{10}^{-3}}\] for the calculation. According to the second law of thermodynamics entropy of the universe always increases for a spontaneous process.