Question

The enthalpy and entropy change for the reaction:
$B{r_2}(l) + C{l_2}(g) \to 2BrCl(g)$
are $30KJmo{l^{ - 1}}$ and $105J{K^{ - 1}}mo{l^{ - 1}}$ respectively. The temperature at which the reaction will be in equilibrium is:
(A) $285.7K$
(B) $273K$
(C) $450K$
(D) $300K$

Answer 1

Enthalpy is the transfer of energy in a reaction and $\vartriangle H$ which is the changing enthalpy. It is like a state function. thought that being a state function means that $\vartriangle H$is independent of the process between initial and final states. Moreover, entropy is the function of the state of the system, So the change in entropy of the system is determined by its initial and final states. Moreover, that process is reversible, the entropy does not change, whereas irreversible processes always increase the total entropy.

Complete step by step answer:
As we are given with the following equation:
$B{r_2}(l) + C{l_2}(g) \to 2BrCl(g)$
Therefore, Enthalpy change that is $\vartriangle H$ is given by $30KJmo{l^{ - 1}} = 30 \times {10^3}J/mol$
an entropy change that is $\vartriangle S$ is $105J{K^{ - 1}}mo{l^{ - 1}}$.
So, we know that,
$\vartriangle G = \vartriangle H - T\vartriangle S$
At equilibrium $\vartriangle H = T\vartriangle S$
$Teq = \dfrac{{\vartriangle H}}{{\vartriangle S}}$
By putting the values of $\vartriangle H$and $\vartriangle S$
Therefore,
$Teq = \dfrac{{30 \times {{10}^3}}}{{105}} \\\ Teq = 285.7K \\\$
At equilibrium, Gibbs free energy change which is $\left( {\vartriangle G^\circ } \right)$ is equal to zero. So, the above equal or the thermodynamic relation is used just to show the relation of $\left( {\vartriangle G^\circ } \right)$ with the enthalpy change$\vartriangle H^\circ$ and the entropy change $\left( {\vartriangle S^\circ } \right)$

So, the correct answer is Option A.

Additional information:
Enthalpy and Entropy changes in a relation which are partly related to each other. The reason for this relationship is that if energy is added to or released from the system, it has to be partitioned into new states. Thus, enthalpy change can also have an effect on Entropy.

Note: Entropy increases as you go from solid to liquid to gas, and you can predict whether entropy change is a positive or negative just by looking at the phases of the reactants and products. Whenever there is an increase in a gas moles, Entropy will increase.