Question

For $A\to B$, $\Delta H=4kcalmo{{l}^{-1}}$, $\Delta S=10calmo{{l}^{-1}}{{K}^{-1}}$. Reaction is spontaneous when temperature is:
A. $400K$
B. $300K$
C. $500K$
D. None of these

Answer 1

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:
For a reaction to be spontaneous the value of $\Delta G$ should be negative. Here, we have asked to find the temperature of the reaction is to be spontaneous.
Also, we have given that
$\Delta H=4kcalmo{{l}^{-1}}$
$\ \Delta S=10calmo{{l}^{-1}}{{K}^{-1}}$
Let us take $\Delta G=0$ for the calculation. Hence we can write the equation as given below;
$\Delta G=\Delta H-T\Delta S$
Or
$\Rightarrow 0=4\times {{10}^{3}}-T\times 10$
$\Rightarrow 10T=4000$
$\Rightarrow T=400K$
For, $\Delta G=0$ the reaction has to occur in $400K$. Hence, it is very clear that above $400K$ the value of $\Delta G$ is equal to a negative value as it is a spontaneous reaction.
So, from the given options only $500K$ is above than $400K$. So, $A\to B$, $\Delta S= 10calmo{{l}^{-1}}{{K}^{-1}}$ reaction is spontaneous when temperature is $500K$.

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.