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Question: What is the equation for finding the equilibrium constant for a chemical reaction?...

What is the equation for finding the equilibrium constant for a chemical reaction?

Explanation

Solution

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:
Kc=[A]a×[B]b×.....[C]c×[D]d×.....{K_c} = \dfrac{{{{\left[ A \right]}^a} \times {{\left[ B \right]}^b} \times .....}}{{{{\left[ C \right]}^c} \times {{\left[ D \right]}^d} \times .....}}
Where, Kc{K_c} is the equilibrium constant, A,B,....A,B,.... is the reactants, C,D,....C,D,.... is the product [A]\left[ A \right] is the equilibrium concentration of AA in moles, aa number of moles of AA .

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 KC{K_C} or KpKp 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:
CO2(g)+H2(g)CO+H2O(g)C{O_2}(g) + {H_2}(g) \rightleftharpoons CO + {H_2}O(g)
Here we are taking 0.1908  moles0.1908\;moles of CO2C{O_2} , 0.0908  moles0.0908\;moles of H2{H_2} , 0.0092  moles0.0092\;moles of COCO and 0.0092  moles0.0092\;moles of H2O{H_2}O vapour were present in the 2.00  L2.00\;L reaction vessel present at equilibrium.
Then, from the equilibrium constant expression:
Kc=[CO][H2O][CO2][H2]{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.
[CO2]=0.1908  mol\left[ {C{O_2}} \right] = 0.1908\;mol
0.0954M\Rightarrow 0.0954M
[H2]=0.0454M[{H_2}] = 0.0454M
[CO]=0.0046M\left[ {CO} \right] = 0.0046M
[H2O]=0.0046M\left[ {{H_2}O} \right] = 0.0046M
On substituting these values in the equilibrium constant formula we get:
Kc=0.0046×0.00460.0454×0.0954{K_c} = \dfrac{{0.0046 \times 0.0046}}{{0.0454 \times 0.0954}}
Kc=0.0049{K_c} = 0.0049
Thus, equilibrium constant Kc{K_c} is 0.00490.0049 . Hence to find the equilibrium constant, we use the following expression:
Kc=[A]a×[B]b×.....[C]c×[D]d×.....{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.