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3 years ago

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Chemistry

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denpristay [2]3 years ago

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Student C: The total mass before and after the reaction is the same.

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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f

Tasya [4]

Answer : The value of $K_c$ for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)$

The expression of $K_c$ will be,

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

First we have to calculate the concentration of $Br_2,Cl_2\text{ and }BrCl$ .

$\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M$

Now we have to calculate the value of $K_c$ for the given reaction.

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

$K_c=\frac{(0.6)^2}{(1)\times (1)}$

$K_c=0.36$

Therefore, the value of $K_c$ for the given reaction is, 0.36

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