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4. The initial concentration of HBr in a reaction vessel is 1.2 mmol/L. If the vessel...

4. The initial concentration of HBr in a reaction vessel is 1.2 mmol/L. If the vessel is heated to 500 K, what is the percentage decomposition of HBr? What is the equilibrium composition of the mixture? 2HBr(g) H2(g) + Br2(g) Kc = 7.7 x 10-11

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Answer #1

Kc = 7.7*10^-11

[HBr] = 1.2*10^-3 M

[H2] = [Br2] = 0

equilibrium

[HBr] = 1.2*10^-3 -2x

[H2] = [Br2] = 0 + x

then

K = H2*Br/(HBr^2)

7.7*10^-11 = (x*x)/(1.2*10^-3 -2x)^2

sqrt(7.7*10^-11) = x/(1.2*10^-3 -2x)

0.00000877496 ((1.2*10^-3 -2x)) = x

113960x = 1.2*10^-3 -2x

113962x = 1.2*10^-3

x = 1.2*10^-3 / (113962) = 1.05298 *10^-8

[HBr] = 1.2*10^-3 -2*1.05298 *10^-8 = 0.00119997894 M

[H2] = [Br2] = 0 + x = 1.05298 *10^-8 M

% decompoisiton = 100 - [HBR] final / [HBr] initial * 100 = 100 - 0.00119997894 /(1.2*10^-3) *100 = 0.001755 %

in equilibrium

[HBr] = 0.00119997894 M

[H2] = [Br2] = 1.05298 *10^-8 M

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