Question

The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant.

            H2(g) + Br2(g) ⇌ 2 HBr(g)   Kc = 3.8 × 104

2 HBr(g) ⇌ H2(g) + Br2(g)         Kc = ?

		2.6 x 10-5
		1.6 × 103
		6.9 × 10-10
		5.9 × 10

Answer 1

rate constant is 2.63 ×10^-5

Ro e eu ivem 3.3 X 10 eand Reaction ke CHG] 4 3.3 10 -ke = 263 x10

Answer 2

Ans :- 2.63 x 10^-5

Explanation :-

Given equilibrium reaction is :

H₂ (g) + Br₂ (g) <-----------> 2 HBr (g) , K_c = 3.8 x 10⁴

Here, K_c is equilibrium constant = [Products]^{stoichiometric coefficient} / Reactants]^{stoichiometric coefficient} .

Expression of K_c for this reaction is :

K_c = [HBr (g)]² / [H₂ (g)].[Br₂ (g)] ........................(1)

Also,

2 HBr (g) <---------------> H₂ (g) + Br₂ (g)

Expression of Equilibrium constant (say K_c^' ) for this reaction is :

K_c' = [H₂ (g)].[Br₂ (g)] / [HBr (g)]² ..............(2)

On comparing (1) and (2) equations , we have the relationship between K_c and K_c' is :

K_c' = 1 / K_c

K_c' = 1 / (3.8 x 10⁴ )

K_c' = 2.63 x 10^-5

Hence, K_c' = 2.63 x 10^-5