Question

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

Answer 1

Hint: Enthalpy is defined as the heat content of a system at constant pressure and is denoted by ‘H’.
Entropy is a thermodynamic state quantity which is a measure of randomness or disorder of the molecules of a system and is denoted by ‘S’.
And Gibbs free energy relates the above 2 quantities as per the following equation:
$\vartriangle G = \vartriangle H - T\vartriangle S$

Complete step by step answer:
Given data:
Enthalpy (H): 30 KJ Mol-1 = 30x103 J Mol-1
Entropy (S): 105 JK-1 Mol-1
Now, in the given question they have asked us to find temperature at equilibrium, which means the Gibbs free energy will be ‘0’.
As per Gibbs free energy equation:
$\vartriangle G = \vartriangle H - T\vartriangle S$
Substituting the values of Enthalpy, Entropy and Gibbs free energy we get
\[0 = 30{\text{ }}x{\text{ }}{10^3}\left( {J{\text{ }}mo{l^{ - 1}}} \right){\text{ }}-{\text{ }}T{\text{ }}x{\text{ }}105{\text{ }}\left( {J{K^{ - 1}}} \right){\text{ }}mo{l^{ - 1}}\]
Therefore,
$T = \dfrac{{30X{{10}^3}}}{{105}}K = 285.71K$
Therefore the correct option is (a).

Additional information:
The Gibbs free energy of the system is defined as “The thermodynamic quantity of the system, the decrease in whose value during a process is equal to useful work done by the system.”
The motive for deriving this equation was to relate enthalpy and entropy in such a way that it arrives at a single function whose sign will determine whether the reaction is spontaneous or not.

Note: For spontaneous function ‘G’ will be negative, and for non-spontaneous function ‘G’ will be positive and it will be zero for equilibrium reactions.
Not only H and S but temperature also is a determining factor for spontaneity as it tells whether the reaction is exothermic or endothermic.