The equilibrium constant for the reaction 2IBr<->I2+Br2 is 8.5×10^-3 at 150°C. If 0.0300 mol of IBr is introduced to a 1L container, what is the concentration of the substance after equilibrium is established?

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The equilibrium constant for the reaction 2IBr<->I2+Br2 is 8.5×10^-3 at 150°C. If 0.0300 mol of IBr is introduced to a 1L container, what is the concentration of the substance after equilibrium is established?

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Answer 1 · 2017-12-11T13:02:19

The is an equilibrium problem, about low-moderate difficulty. An $ICE$ $"ICE"$ table is needed to efficiently solve this. Concentration calculations I will omit so my answer is more clear, however, be meticulous with these on exams, I can't count how many stupid mistakes I've made over these.

$2 I B r (g) ⇌ I_{2} (g) + B r_{2} (g)$ $2IBr(g) rightleftharpoons I_2(g) + Br_2(g)$ where $K_{c} = 8.5 \cdot 10^{- 3}$ $K_c = 8.5*10^-3$ , and

$K_{c} = \frac{[I_{2}] [B r_{2}]}{{[I B r]}^{2}}$ $K_c = ([I_2][Br_2])/([IBr]^2)$

Hence,

$2 I B r (g) ⇌ I_{2} (g) + B r_{2} (g)$ $2IBr(g) rightleftharpoons I_2(g) + Br_2(g)$
puu.sh

$K_{c} = 8.5 \cdot 10^{- 3} = \frac{x^{2}}{{(0.03 - x)}^{2}}$ $K_c = 8.5*10^-3 = (x^2)/(0.03-x)^2$
$K_{c} = 9.2 \cdot 10^{- 2} = \frac{x}{0.03 - x}$ $K_c = 9.2*10^-2 = x/(0.03 -x)$
$therefore x approx 0.0025$

The concentrations at equilibrium (which I've calculated per the $"ICE"$ table) are,

$[IBr] = 0.0275M$
$[I_2] = [Br_2] = 0.0025M$

This is reasonable because $K_c < 1$ , and the reaction won't be very active in the right direction, hence the small "product" concentrations.

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The equilibrium constant for the reaction 2IBr<->I2+Br2 is 8.5×10^-3 at 150°C. If 0.0300 mol of IBr is introduced to a 1L container, what is the concentration of the substance after equilibrium is established?

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