Explain the calculation of equilibrium concentrations given the value of the equilibrium constant.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) To calculate equilibrium concentrations when initial concentrations are known and the equilibrium constant $(K_c)$ is provided,follow these steps:
Step-$1$: Write the balanced chemical equation for the reaction.
Step-$2$: Create an $ICE$ (Initial,Change,Equilibrium) table under the balanced equation for each substance involved.
$(i)$ List the initial concentration.
$(ii)$ Define the change in concentration as $x$ (in $mol/L$) based on the stoichiometry of the reaction.
$(iii)$ Express the equilibrium concentration as the sum of the initial concentration and the change.
Step-$3$: Substitute these equilibrium concentrations into the equilibrium constant expression $(K_c)$ and solve the resulting algebraic equation for $x$. If a quadratic equation is obtained,select the root that is physically meaningful (i.e.,concentrations must be positive).
Step-$4$: Use the calculated value of $x$ to determine the actual equilibrium concentrations of all reactants and products.
Step-$5$: Verify the results by substituting the calculated equilibrium concentrations back into the $K_c$ expression to ensure they yield the given constant.

Explore More

Similar Questions

For the reaction,$A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$ at $25 \, ^oC$,$2 \, moles$ of $A$,$1 \, mole$ of $B$ and $1 \, mole$ of $C$ are present in a $1 \, L$ vessel. If $K_c$ for the reaction is $2$,then the reaction will proceed in:

$PCl_5 \rightleftharpoons PCl_3 + Cl_2$. If the equilibrium constant $(K_C)$ for the above reaction at $500 \ K$ is $1.79$ and the equilibrium concentrations of $PCl_5$ and $PCl_3$ are $1.41 \ M$ and $1.59 \ M$,respectively,then the concentration of $Cl_2$ is approximately: (in $M$)

The unit of equilibrium constant $K$ for the reaction $A + B \rightleftharpoons C$ would be

The value of $K_{c}$ for the reaction $2 A \rightleftharpoons B + C$ is $2 \times 10^{-3}$. At a given time,the composition of the reaction mixture is $[ A ] = [ B ] = [ C ] = 3 \times 10^{-4} \ M$. In which direction will the reaction proceed?

An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo