Question

Standard entropy of ${X_2},{Y_2}$ and $X{Y_3}$are $60,40$ and $50ZJ{K^{ - 1}}mo{l^{ - 1}}$ respectively. For the reaction
$\dfrac{1}{2}{X_2} + \dfrac{3}{2}{Y_2} \to X{Y_3}$ , $\Delta H = - 30kJ$ , to be at equilibrium , the temperature will be:
A. $1250K$
B. $500K$
C. $750K$
D. $1000K$

Answer 1

Hint:Use $\Delta {S_{reaction}} = \sum {\Delta {S_{product}}} - \sum {\Delta {S_{reac\tan t}}} $by which we will get standard entropy of the reaction. After getting the standard entropy of the reaction use the equation $\Delta G = \Delta H - T\Delta S$( $\Delta G = 0$at equilibrium).
Formula Used:
$\Delta G = \Delta H - T\Delta S$
Where, $\Delta G$
$\Delta S$ = Entropy of reaction
$T$ = Temperature
= Free energy
$\Delta H$ = Change in heat

Complete step by step answer:
Given reaction is,
$\dfrac{1}{2}{X_2} + \dfrac{3}{2}{Y_2} \to X{Y_3}$
And for the above reaction:
 $\Delta H = - 30kJ$
Now, According to the question:
Standard Entropy for ${X_2} = 60J{K^{ - 1}}mo{l^{ - 1}}$
                                       ${Y_2} = 40J{K^{ - 1}}mo{l^{ - 1}}$
                                      $X{Y_3} = 50J{K^{ - 1}}mo{l^{ - 1}}$
So, $\Delta {S_{reac\tan t}} = (1 \times 60 + 3 \times 40)J{K^{ - 1}}mo{l^{ - 1}}$
$ \Rightarrow 60 + 120$
$ \Rightarrow 180J{K^{ - 1}}mo{l^{ - 1}}$
And ,
$\Delta {S_{product}} = 2 \times 50J{K^{ - 1}}mo{l^{ - 1}}$
$ \Rightarrow 100J{K^{ - 1}}mo{l^{ - 1}}$
Now, standard entropy for the reaction will be:
$\Delta {S_{reaction}} = \Delta {S_{product}} - \Delta {S_{reac\tan t}}$
$ \Rightarrow (100 - 180)J{K^{ - 1}}mo{l^{ - 1}}$
$ \Rightarrow - 80J{K^{ - 1}}mo{l^{ - 1}}$
Now, As from the equation:
$\Delta G = \Delta H - T\Delta S$
At equilibrium, $\Delta G = 0$
So,
$\Delta H - T\Delta S = 0$
$ \Rightarrow \Delta H = T\Delta S$
From the question,
$\Delta H = 1000 \times - (60)$
$ \Rightarrow 1000 \times ( - 60) = T \times ( - 80)$
$ \Rightarrow T = 750K$

Hence, option C is correct.

Additional Information:
Thermodynamics is a major topic in chemistry. There are two definitions of entropy which are mostly used: 1. The thermodynamic definition and 2. The statistical mechanics definition.
In classical thermodynamics, the details of a system are not considered as microscopically but the properties of a system are defined by the thermodynamic variable. Such as temperature, pressure, entropy, and heat capacity.
The statistical definition of entropy and other thermodynamic properties were proposed later. According to this, thermodynamic properties are described in terms of the statistics of the motions of a system ( microscopically )– modeled at first classically, for example Newtonian particles constituting a gas, and later quantum-mechanically (photons, phonons, spins and so on.).

Note:
The calorific value can be defined as the amount of heat produced on combusting a unit volume of gas and can be expressed in $kcal/{m^3}$, $kJ/{m^3}$ . Calorific value depends directly on the methane content of $LFG$ , such that the higher the methane content, the greater the calorific value.