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Question: The value of \({\log _{10}}K\) for a reaction \(A \rightleftharpoons B\) is_____________. Given \(...

The value of log⁡10K{\log _{10}}K for a reaction A⇌BA \rightleftharpoons B is_____________.
Given ΔrH298K0=−54.07J/mol,ΔrS298K0=10JK−1mol−1,R=8.314JK−1mol−1\Delta rH_{298K}^0 = - 54.07J/mol,{\Delta _r}S_{298K}^0 = 10J{K^{ - 1}}mo{l^{ - 1}},R = 8.314J{K^{ - 1}}mo{l^{ - 1}}.
A. 5
B. 10
C. 95
D. 100

Explanation

Solution

Hint: First, we will apply the formulas of Gibb’s energy and then simply equate them. Then substitute the values into the calculated equation to get to our required answer. Refer to the solution below.

Complete step-by-step answer:
As per given in the questions, we have to find out the equilibrium constant of the equation (log⁡10K{\log _{10}}K means the equilibrium constant).
As we know, the formula of Gibb’s energy is given as-
⇒ΔG0=ΔH0−TΔS0\Rightarrow \Delta {G^0} = \Delta {H^0} - T\Delta {S^0}
As well as we also know that-
⇒ΔG0=−2.303RTlog⁡K\Rightarrow \Delta {G^0} = - 2.303RT\log K
Now, equating both the equations given above we will have-
⇒ΔH0−TΔS0=−2.303RTlog⁡K  ⇒−(−ΔH0+TΔS0)=−2.303RTlog⁡K  ⇒TΔS0−ΔH0=2.303RTlog⁡K  ⇒log⁡K=TΔS0−ΔH02.303RT  \Rightarrow \Delta {H^0} - T\Delta {S^0} = - 2.303RT\log K \\\ \\\ \Rightarrow - \left( { - \Delta {H^0} + T\Delta {S^0}} \right) = - 2.303RT\log K \\\ \\\ \Rightarrow T\Delta {S^0} - \Delta {H^0} = 2.303RT\log K \\\ \\\ \Rightarrow \log K = \dfrac{{T\Delta {S^0} - \Delta {H^0}}}{{2.303RT}} \\\
It was given by the question that the value of ΔH0\Delta {H^0} is −54.07J/mol - 54.07J/mol.
The value of ΔS0\Delta {S^0} as per given by the question is 10JK−1mol−110J{K^{ - 1}}mo{l^{ - 1}}.
And the value of R as per given by the question is 8.314JK−1mol−18.314J{K^{ - 1}}mo{l^{ - 1}}.
Now, substituting the given values into the given equation, we have-
⇒log⁡K=TΔS0−ΔH02.303RT  ⇒log⁡K=298×10J−(−54.07×103)2.303×8.314×298  ⇒log⁡K=−570505705  ⇒log⁡K=10  \Rightarrow \log K = \dfrac{{T\Delta {S^0} - \Delta {H^0}}}{{2.303RT}} \\\ \\\ \Rightarrow \log K = \dfrac{{298 \times 10J - \left( { - 54.07 \times {{10}^3}} \right)}}{{2.303 \times 8.314 \times 298}} \\\ \\\ \Rightarrow \log K = - \dfrac{{57050}}{{5705}} \\\ \\\ \Rightarrow \log K = 10 \\\
Therefore, log⁡10K=10{\log _{10}}K = 10
Hence, it can be said that option B is the correct option.

Note: The equilibrium constant (typically denoted as symbol K) of a chemical reaction provides an insight into the relationship between products and reactants in the equilibrium of a chemical reaction. The equilibrium constant can be defined for a chemical reaction as the ratio of the volume of reactant to the quantity of product to determine the chemical behaviour.