Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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/litB. ${{\left( mol/lit \right)}^{-1}}$C. $li{{t}^{2}}mo{{l}^{-2}}$D. ${{k}_{c}}$ has no units

Last updated date: 13th Jun 2024
Total views: 393.9k
Views today: 11.93k
Verified
393.9k+ 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 ${{\left( mol/lit \right)}^{-1}}$.

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 $\left( {{k}_{c}} \right)$.