
The value of ${\log _{10}}K$ for a reaction $A \rightleftharpoons B$ is_____________.
Given $\Delta rH_{298K}^0 = - 54.07J/mol,{\Delta _r}S_{298K}^0 = 10J{K^{ - 1}}mo{l^{ - 1}},R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$.
A. 5
B. 10
C. 95
D. 100
Answer
597.6k+ views
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 (${\log _{10}}K$ means the equilibrium constant).
As we know, the formula of Gibb’s energy is given as-
$ \Rightarrow \Delta {G^0} = \Delta {H^0} - T\Delta {S^0}$
As well as we also know that-
$ \Rightarrow \Delta {G^0} = - 2.303RT\log K$
Now, equating both the equations given above we will have-
$
\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 $\Delta {H^0}$ is $ - 54.07J/mol$.
The value of $\Delta {S^0}$ as per given by the question is $10J{K^{ - 1}}mo{l^{ - 1}}$.
And the value of R as per given by the question is $8.314J{K^{ - 1}}mo{l^{ - 1}}$.
Now, substituting the given values into the given equation, we have-
$
\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, ${\log _{10}}K = 10$
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.
Complete step-by-step answer:
As per given in the questions, we have to find out the equilibrium constant of the equation (${\log _{10}}K$ means the equilibrium constant).
As we know, the formula of Gibb’s energy is given as-
$ \Rightarrow \Delta {G^0} = \Delta {H^0} - T\Delta {S^0}$
As well as we also know that-
$ \Rightarrow \Delta {G^0} = - 2.303RT\log K$
Now, equating both the equations given above we will have-
$
\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 $\Delta {H^0}$ is $ - 54.07J/mol$.
The value of $\Delta {S^0}$ as per given by the question is $10J{K^{ - 1}}mo{l^{ - 1}}$.
And the value of R as per given by the question is $8.314J{K^{ - 1}}mo{l^{ - 1}}$.
Now, substituting the given values into the given equation, we have-
$
\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, ${\log _{10}}K = 10$
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.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

