Question

The variation of the equilibrium constant with temperature is given below :

Temperature Equilibrium Constant

$$\begin{gathered} T_1={25}^{o}C,K_1=10 \\ {T}_2={100}^{o}C,K_2=100 \end{gathered}$$

The value of $∆H^0,∆G^0$ at $T_1$and $∆G^0$at $T_2$ (in $Kjmo{l}^{-1}$) respectively, are close to

[use $R=8.314J{K}^{-1}mo{l}^{-1}$]

• A. $28.4,-7.14and-5.71$
• B. $0.64,-7.14and-5.71$
• C. $28.4,-5.71and-14.29$
• D. $0.64,-5.71and-14.29$

Answer 1

The correct option is C

$28.4,-5.71and-14.29$

The explanation for the correct answer:-

In the case of option C.

Step 1. Given data:

$For{T}_1=298K,K_1=10$

$For{T}_2=373K,K_2=100$

$R=8.314J{K}^{-1}mo{l}^{-1}$

Step 2. Formula used:

$\log \left(\frac{K_2}{K_1}\right)=\frac{∆H}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Where $K,∆H,R,T$ are equilibrium constant, enthalpy change, gas constant, and temperature respectively.

Step 3. Calculations:

For $∆H$ calculation as

$$\begin{gathered} \log \left(\frac{K_2}{K_1}\right)=\frac{∆H}{2.303R}(\frac{1}{T_1}-\frac{1}{T_2}) \\ \log \left(\frac{100}{10}\right)=\frac{∆H}{2.303\left({\displaystyle \frac{8.314}{1000}}\right)}(\frac{1}{298}-\frac{1}{373}) \\ ∆H=28.4KJ/mole \end{gathered}$$

Now, standard Gibbs free energy change =$∆G$

For $∆{G^0}_{1}and∆{G^0}_2$calculation as

$$\begin{gathered} ∆{G^0}_1=-2.303RT\log K \\ ∆{G^0}_1=-2.303\times \frac{8.314}{1000}\times 298\times \log 10 \\ ∆{G^0}_1=-5.71KJ/mole \\ ∆{G^0}_2=-2.303\times \frac{8.314}{1000}\times 373\times \log 100 \\ ∆{G^0}_2=-14.29KJ/mole \end{gathered}$$

Hence the correct answer is c

The explanation for the incorrect answer:-

Option (A) $28.4,-7.14and-5.71$

1. The equilibrium constant's value falls as temperature rises.
2. An rise in temperature increases the value of the equilibrium constant when the forward reaction is endothermic
3. Option a is incorrect

Option (B) $0.64,-7.14and-5.71$

1. As the temperature fluctuates, so does the equilibrium position.
2. A figure that illustrates the relationship between the quantities of reactants and products in an equilibrium chemical reaction at a specific temperature.
3. Hence Option b is incorrect

Option (D) $0.64,-5.71and-14.29$

1. The ratio of the forward reaction's velocity constant to the backward reaction's velocity constant is known as the equilibrium constant." also known as
2. "It is the ratio of product to reactant molar concentration to product molar concentration.
3. Hence option d is incorrect

Therefore, option (C) is the correct answer