Question

Give the reaction between 2 gases represented by ${{A}_{2}}$ and \[{{B}_{2}}\] to give the compound \[A{{B}_{(g)}}\]
${{A}_{2(g)}}+{{B}_{2(g)}}\rightleftarrows 2A{{B}_{(g)}}$
At equilibrium, the concentration
Of ${{A}_{2}}$= $3.0\times {{10}^{-3}}M$
Of \[{{B}_{2}}\]= $4.2\times {{10}^{-3}}M$
Of AB = $2.8\times {{10}^{-3}}M$
If the reaction takes place in sealed vessel at 527º C, then the value of ${{K}_{c}}$will be:
A. 1.9
B. 0.62
C. 4.5
D. 2.0

Answer 1

Hint: The equilibrium constant is the value of the reaction quotient that is calculated from the expression for chemical equilibrium. It depends on the ionic strength and temperature and is independent of the concentrations of reactants and products in a solution.

Complete Step by step solution: The equilibrium constant depends on whether the chemical reaction involves homogeneous equilibrium or heterogeneous equilibrium. All of the products and reactants are in the same phase for a reaction are known to be at homogeneous equilibrium. More than one phase is present for reactions that reach heterogeneous equilibrium. For any given temperature, there is only one value for the equilibrium constant. ${{K}_{c}}$ only changes if the temperature at which the reaction occurs changes. You can make some predictions about the chemical reaction based on whether the equilibrium constant is large or small. The equilibrium reaction is as given below
${{A}_{2(g)}}+{{B}_{2(g)}}\rightleftarrows 2A{{B}_{(g)}}$
Expression for the equilibrium constant is shown as:
${{K}_{c}}=\dfrac{{{[AB]}^{2}}}{[{{A}_{2}}][{{B}_{2}}]}$
By substituting the values in the equation
${{K}_{c}}=\dfrac{{{(2.8\times {{10}^{-3}})}^{2}}}{3\times {{10}^{-3}}\times 4.2\times {{10}^{-3}}}$= 0.62

Hence option B is the correct answer.

Note: If the value of ${{K}_{c}}$ is very large then the equilibrium favors the reaction to the right direction and there are more products than reactants. The reaction is said to be quantitative and if the value for the equilibrium constant is small, then the equilibrium favors the reaction to the left, and there are more reactants than products. If the value of ${{K}_{c}}$approaches zero the reaction may be considered not to occur.