Question

At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g) At equilibrium the concentration of Br2 was measured to be 0.0955 M. What is Kc for this reaction at this temperature

Answer 1

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of  $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

                            $2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc.              0.500 M              0  M        0 M  

At eqm. conc.            (0.500-2x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x =  0.0955 M

Now put all the given values in this expression, we get :

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029