
What are the units of equilibrium constant $\left( {{k}_{c}} \right)$ for the following reaction?
${{C}_{\left( s \right)}}+{{O}_{2}}_{\left( g \right)}\rightleftharpoons C{{O}_{2}}\left( g \right)$
A. mol/lit
B. ${{\left( mol/lit \right)}^{-1}}$
C. $li{{t}^{2}}mo{{l}^{-2}}$
D. ${{k}_{c}}$ has no units
Answer
555k+ views
Hint: Equilibrium constant is denoted by the symbol $\left( {{k}_{c}} \right)$. It is found that the expression for equilibrium constant $\left( {{k}_{c}} \right)$ is given by the equation: ${{k}_{c}}=\dfrac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}$
Complete Step by step solution:
- Equilibrium constant helps us to determine whether the reaction will have a higher concentration of reactants or products.
- We are being provided with the equation:
\[C\left( s \right)+{{O}_{2\left( g \right)}}\rightleftharpoons C{{O}_{2}}\left( g \right)\]
- As we know that equilibrium constant $\left( {{k}_{c}} \right)$ is given by the equation
${{k}_{c}}=\dfrac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}$
So, we can write equilibrium constant for the given reaction as:
\[{{k}_{c}}=\dfrac{\left[ C{{O}_{2}} \right]}{\left[ C \right]\left[ {{O}_{2}} \right]}\]
As we know that the unit of concentration is mol/litre. Now, putting the value of units in the above formula we get:
\[\begin{align}
& {{k}_{c}}=\dfrac{mol\text{ }li{{t}^{-1}}}{\left( mol\text{ }li{{t}^{-1}} \right)\left( mol\text{ }li{{t}^{-1}} \right)} \\
& \implies\dfrac{1}{mol\text{ }li{{t}^{-1}}} \\
& \implies\left( \dfrac{mo{{l}^{-1}}}{li{{t}^{-1}}} \right) \\
& \implies{{\left( \dfrac{mol}{lit} \right)}^{-1}} \\
\end{align}\]
- Hence, we can conclude that the correct option is (b) that is the units of equilibrium constant $\left( {{k}_{c}} \right)$ for the given reaction is ${{\left( mol/lit \right)}^{-1}}$.
Additional information:
- It is also found that knowledge of equilibrium constant is necessary for understanding various biochemical processes like acid-base homeostasis and oxygen transport by haemoglobin in blood. At equilibrium, various known equilibrium constant values can be used to determine the composition of the system.
- As we know that the equilibrium constant can be used to predict the extent of reaction, direction of reaction.
Note: - It is found that the equilibrium constant $\left( {{k}_{c}} \right)$ depends upon temperature. In case of exothermic reactions increasing the temperature will reduce , and in endothermic reactions increasing the temperature will increase $\left( {{k}_{c}} \right)$.
- And change in concentration, catalyst, pressure etc. have no effect on equilibrium constant.
Complete Step by step solution:
- Equilibrium constant helps us to determine whether the reaction will have a higher concentration of reactants or products.
- We are being provided with the equation:
\[C\left( s \right)+{{O}_{2\left( g \right)}}\rightleftharpoons C{{O}_{2}}\left( g \right)\]
- As we know that equilibrium constant $\left( {{k}_{c}} \right)$ is given by the equation
${{k}_{c}}=\dfrac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reactant}$
So, we can write equilibrium constant for the given reaction as:
\[{{k}_{c}}=\dfrac{\left[ C{{O}_{2}} \right]}{\left[ C \right]\left[ {{O}_{2}} \right]}\]
As we know that the unit of concentration is mol/litre. Now, putting the value of units in the above formula we get:
\[\begin{align}
& {{k}_{c}}=\dfrac{mol\text{ }li{{t}^{-1}}}{\left( mol\text{ }li{{t}^{-1}} \right)\left( mol\text{ }li{{t}^{-1}} \right)} \\
& \implies\dfrac{1}{mol\text{ }li{{t}^{-1}}} \\
& \implies\left( \dfrac{mo{{l}^{-1}}}{li{{t}^{-1}}} \right) \\
& \implies{{\left( \dfrac{mol}{lit} \right)}^{-1}} \\
\end{align}\]
- Hence, we can conclude that the correct option is (b) that is the units of equilibrium constant $\left( {{k}_{c}} \right)$ for the given reaction is ${{\left( mol/lit \right)}^{-1}}$.
Additional information:
- It is also found that knowledge of equilibrium constant is necessary for understanding various biochemical processes like acid-base homeostasis and oxygen transport by haemoglobin in blood. At equilibrium, various known equilibrium constant values can be used to determine the composition of the system.
- As we know that the equilibrium constant can be used to predict the extent of reaction, direction of reaction.
Note: - It is found that the equilibrium constant $\left( {{k}_{c}} \right)$ depends upon temperature. In case of exothermic reactions increasing the temperature will reduce , and in endothermic reactions increasing the temperature will increase $\left( {{k}_{c}} \right)$.
- And change in concentration, catalyst, pressure etc. have no effect on equilibrium constant.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which type of resource is iron ore A Renewable B Biotic class 11 social science CBSE

10 examples of friction in our daily life

Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

