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Question: Standard entropy of \({X_2},{Y_2}\) and \(X{Y_3}\)are \(60,40\) and \(50ZJ{K^{ - 1}}mo{l^{ - 1}}\) r...

Standard entropy of X2,Y2{X_2},{Y_2} and XY3X{Y_3}are 60,4060,40 and 50ZJK−1mol−150ZJ{K^{ - 1}}mo{l^{ - 1}} respectively. For the reaction
12X2+32Y2→XY3\dfrac{1}{2}{X_2} + \dfrac{3}{2}{Y_2} \to X{Y_3} , ΔH=−30kJ\Delta H = - 30kJ , to be at equilibrium , the temperature will be:
A. 1250K1250K
B. 500K500K
C. 750K750K
D. 1000K1000K

Explanation

Solution

Use ΔSreaction=∑ΔSproduct−∑ΔSreactan⁡t\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 ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S( ΔG=0\Delta G = 0at equilibrium).
Formula Used:
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
Where, ΔG\Delta G
ΔS\Delta S = Entropy of reaction
TT = Temperature
= Free energy
ΔH\Delta H = Change in heat

Complete step by step answer:
Given reaction is,
12X2+32Y2→XY3\dfrac{1}{2}{X_2} + \dfrac{3}{2}{Y_2} \to X{Y_3}
And for the above reaction:
ΔH=−30kJ\Delta H = - 30kJ
Now, According to the question:
Standard Entropy for X2=60JK−1mol−1{X_2} = 60J{K^{ - 1}}mo{l^{ - 1}}
Y2=40JK−1mol−1{Y_2} = 40J{K^{ - 1}}mo{l^{ - 1}}
XY3=50JK−1mol−1X{Y_3} = 50J{K^{ - 1}}mo{l^{ - 1}}
So, ΔSreactan⁡t=(1×60+3×40)JK−1mol−1\Delta {S_{reac\tan t}} = (1 \times 60 + 3 \times 40)J{K^{ - 1}}mo{l^{ - 1}}
⇒60+120\Rightarrow 60 + 120
⇒180JK−1mol−1\Rightarrow 180J{K^{ - 1}}mo{l^{ - 1}}
And ,
ΔSproduct=2×50JK−1mol−1\Delta {S_{product}} = 2 \times 50J{K^{ - 1}}mo{l^{ - 1}}
⇒100JK−1mol−1\Rightarrow 100J{K^{ - 1}}mo{l^{ - 1}}
Now, standard entropy for the reaction will be:
ΔSreaction=ΔSproduct−ΔSreactan⁡t\Delta {S_{reaction}} = \Delta {S_{product}} - \Delta {S_{reac\tan t}}
⇒(100−180)JK−1mol−1\Rightarrow (100 - 180)J{K^{ - 1}}mo{l^{ - 1}}
⇒−80JK−1mol−1\Rightarrow - 80J{K^{ - 1}}mo{l^{ - 1}}
Now, As from the equation:
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S
At equilibrium, ΔG=0\Delta G = 0
So,
ΔH−TΔS=0\Delta H - T\Delta S = 0
⇒ΔH=TΔS\Rightarrow \Delta H = T\Delta S
From the question,
ΔH=1000×−(60)\Delta H = 1000 \times - (60)
⇒1000×(−60)=T×(−80)\Rightarrow 1000 \times ( - 60) = T \times ( - 80)
⇒T=750K\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/m3kcal/{m^3}, kJ/m3kJ/{m^3} . Calorific value depends directly on the methane content of LFGLFG , such that the higher the methane content, the greater the calorific value.