Question

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 1

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:

& {{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{aligned}$$ **\- 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.