What is the equation for finding the equilibrium constant for a chemical reaction?
Answer
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Hint: Equilibrium constant is the relationship between reactants and products of a reaction at equilibrium with respect to a specific unit. By using equilibrium constant expressions and the calculations involved with both the concentration and the partial pressure equilibrium constant.
Equilibrium constant expression is equal to:
$ {K_c} = \dfrac{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times .....}}{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times .....}} $
Where, $ {K_c} $ is the equilibrium constant, $ A,B,.... $ is the reactants, $ C,D,.... $ is the product $ \left[ A \right] $ is the equilibrium concentration of $ A $ in moles, $ a $ number of moles of $ A $ .
Complete Step By Step Answer:
We have to do two things for the calculation of the equilibrium constant.
First, we will write the balanced equation for the chemical reaction including the physical states of the species. We will find the value of the $ {K_C} $ or $ Kp $ from the equilibrium expression. Thus, after substituting these values in the equilibrium expression, we will get the value of equilibrium constant.
For example, for the chemical equation:
$ C{O_2}(g) + {H_2}(g) \rightleftharpoons CO + {H_2}O(g) $
Here we are taking $ 0.1908\;moles $ of $ C{O_2} $ , $ 0.0908\;moles $ of $ {H_2} $ , $ 0.0092\;moles $ of $ CO $ and $ 0.0092\;moles $ of $ {H_2}O $ vapour were present in the $ 2.00\;L $ reaction vessel present at equilibrium.
Then, from the equilibrium constant expression:
$ {K_c} = \dfrac{{[CO][{H_2}O]}}{{[C{O_2}][{H_2}]}} $
As this is not in moles per litre so we will change them in moles per litre.
$ \left[ {C{O_2}} \right] = 0.1908\;mol $
$ \Rightarrow 0.0954M $
$ [{H_2}] = 0.0454M $
$ \left[ {CO} \right] = 0.0046M $
$ \left[ {{H_2}O} \right] = 0.0046M $
On substituting these values in the equilibrium constant formula we get:
$ {K_c} = \dfrac{{0.0046 \times 0.0046}}{{0.0454 \times 0.0954}} $
$ {K_c} = 0.0049 $
Thus, equilibrium constant $ {K_c} $ is $ 0.0049 $ . Hence to find the equilibrium constant, we use the following expression:
$ {K_c} = \dfrac{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times .....}}{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times .....}} $
Note:
With the help of the equilibrium constant, we are able to understand if the reaction tends to have a higher concentration of products or reactants at equilibrium. The state in which the rate of the forward reaction equals the rate of the reverse reaction is known as equilibrium.
Equilibrium constant expression is equal to:
$ {K_c} = \dfrac{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times .....}}{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times .....}} $
Where, $ {K_c} $ is the equilibrium constant, $ A,B,.... $ is the reactants, $ C,D,.... $ is the product $ \left[ A \right] $ is the equilibrium concentration of $ A $ in moles, $ a $ number of moles of $ A $ .
Complete Step By Step Answer:
We have to do two things for the calculation of the equilibrium constant.
First, we will write the balanced equation for the chemical reaction including the physical states of the species. We will find the value of the $ {K_C} $ or $ Kp $ from the equilibrium expression. Thus, after substituting these values in the equilibrium expression, we will get the value of equilibrium constant.
For example, for the chemical equation:
$ C{O_2}(g) + {H_2}(g) \rightleftharpoons CO + {H_2}O(g) $
Here we are taking $ 0.1908\;moles $ of $ C{O_2} $ , $ 0.0908\;moles $ of $ {H_2} $ , $ 0.0092\;moles $ of $ CO $ and $ 0.0092\;moles $ of $ {H_2}O $ vapour were present in the $ 2.00\;L $ reaction vessel present at equilibrium.
Then, from the equilibrium constant expression:
$ {K_c} = \dfrac{{[CO][{H_2}O]}}{{[C{O_2}][{H_2}]}} $
As this is not in moles per litre so we will change them in moles per litre.
$ \left[ {C{O_2}} \right] = 0.1908\;mol $
$ \Rightarrow 0.0954M $
$ [{H_2}] = 0.0454M $
$ \left[ {CO} \right] = 0.0046M $
$ \left[ {{H_2}O} \right] = 0.0046M $
On substituting these values in the equilibrium constant formula we get:
$ {K_c} = \dfrac{{0.0046 \times 0.0046}}{{0.0454 \times 0.0954}} $
$ {K_c} = 0.0049 $
Thus, equilibrium constant $ {K_c} $ is $ 0.0049 $ . Hence to find the equilibrium constant, we use the following expression:
$ {K_c} = \dfrac{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times .....}}{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times .....}} $
Note:
With the help of the equilibrium constant, we are able to understand if the reaction tends to have a higher concentration of products or reactants at equilibrium. The state in which the rate of the forward reaction equals the rate of the reverse reaction is known as equilibrium.
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