Chemistry Forum

Consider the gas-phase reaction

Xe + 2F₂ ⇌ XeF₄

a) Calculate the equilibrium constant from the observation that at some temperature, the extent of reaction is 50% when 0.20 mol of Xe and 0.40 mol of F₂ have been mixed in an empty 1.00-L-bulb.

b)How many moles of F₂ would have to be added to the equilibrium mixuture of part (a) in order to increase the conversion of Xe to XeF₄ to 80%?

\((a)K= \frac{[XeF_4]}{[Xe][F_2]^2}= \frac{(0.10)}{(010)(0.20)^2}=25,\)

(b) 80% of 0.20 moles of Xe= 0.80 × 0.20 =0.16 mol XeF₄

\(\frac{(0.16)}{(0.040)(0.20 + x -0.12)^2}=25\)

x=0.32 mol F₂

0,16/(0,04*(0,2+2x)^2) = 25

Solve for x

But author gave the answer 0.32 mol of F₂ which is correct as per my calculations.

As per your calculation x=0.1 mol of F₂

But where does 0,12 comes from in your calculation.

From ICE table, I found that we have to deduct 0.12 mol from 0.2 mol of F₂ and 0.06 mol from 0.10 mol of Xe. That's why my answer matches with author's answer.

and 0.6 mol from 0.10 mol of Xe.

????

But answer is correct.

That's a typographical mistake. I corrected it in my post.