The reaction will shift to the right in the direction of products. The equilibri

Question

The reaction will shift to the right in the direction of products. The equilibrium constant will decrease. The equilibrium constant will increase. No effect will be observed. The reaction will shift to the left in the direction of reactants. The equilibrium constant is given for one of the reactions below. Determine the value of the missing equilibrium constant. H_2(g) + Br_21(g) = 2 HBr(g) Kc = 3.8 times 10^4 4 HBr(g) = 2 H _2(x) + 2 Br_2(g) Kc = ? 5.1 times 10^-3 1.6 times 10^3 19 times 1064 2.6 times 10^-5 6.9 x 10^-10 Define dynamic equilibrium. the rate of the reverse reaction is faster than the rate of the forward reaction

Explanation / Answer

Q8.

Not that the reaction is atually inverted and m ultiplied by 2

so the K can be related via algebra as follows

K1 = [HBr]^2 /([H2][Br2])

K2 = [H2]^2[Br]^2] /[HBr]^4

Note that

if we power to the second power --> K1 becomes = K1 = [HBr]^4 /([H2]^2[Br2]^2)

Yet, it must be inverted, so K1 becomes now --> /([H2]^2[Br2]^2) / [HBr]^4

which is what we wanted

the overal effect in K

K2 = ( K1)^2 ) ^-1

K2 = 1/K1^2 = 1/(3.8*10^4) ^2

K2 = 6.925*10^-10

best answer is E

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