Question

The enthalpy and entropy change for the reaction, $B{r_2}(l) + C{l_2}(g) \to 2BrCl(g)$ are $7.1Kcal$$mol$ and $25cal{K^{ - 1}}mo{l^{ - 1}}$ respectively. This reaction will be spontaneous at:
(A). $10^\circ C$
(B). $280K$
(C). $290K$
(D). $5^\circ C$

Answer 1

Hint: In this question, enthalpy and entropy changes are given for the reaction. Therefore, enthalpy change is the amount of heat absorbed or evolved in a reaction which is carried out at constant pressure. Change in entropy is defined as the difference between the entropies of the final and initial state.

Complete step by step answer:
In thermodynamics a spontaneous reaction is the reaction that favors the formation of products at the conditions under which the reaction is occurring. And the combination which shows decrease in energy and an increase in entropy means that the combustion reaction occurs spontaneously. Here by using the relation
$\Delta G = \Delta H - T\Delta S$
Here, $\Delta H = 7100$ , $\Delta S = 25T$ (given)
For a reaction to be spontaneous,
$\Delta G < 0$
$ \Rightarrow \Delta G = \Delta H - T\Delta S < 0$
$7100 - 25T < 0$
$T > 284K$
As for a spontaneous reaction, the sign on delta G must be negative. This spontaneous reaction will always occur when Delta H is negative and Delta S is positive, and a reaction will always be non-spontaneous when Delta H is positive and Delta S is negative.
 Option C is the correct one that is $290K$ .

Note:
Delta H is the change in enthalpy and Delta V is the change in internal energy. As internal energy is the amount of energy that a system has.
And if the $\Delta G$ is equal to zero, then there is –ve net charge in A and B , as the system is at equilibrium.
If $\Delta G$is positive, the process is nonspontaneous and it may proceed spontaneously in the reverse direction and if the $\Delta G$ is negative, the process is spontaneous and it may further proceed in the forward direction.