Imat 2019 q43 chem

Practice Question Solving
1

Hi can I please have some help with this question. I tried using Kc= (c)^2 x (D)/ (B)^2 X (A) but I realised that I am not given their concentration . How would I go about this question please ?

Consider the following reversible reaction at temperature T. A + 2B ⇌ 2C + D

At equilibrium, there are 0.5 moles of A, 0.2 moles of B, 0.5 moles of C and 0.8 moles of D, all in a vessel of volume V.

What is the value of the equilibrium constant, Kc, at this temperature?

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2

Hi!
Yup you’re right we are not given concentration we have been given moles of reactants and products and the volume (V).
concentration can be calculated by dividing n with V as c=n/V
Here we can ignore V as it will basically cancel out.
Thus Kc= 0.5^2 * 0.8/0.5 * 0.2^2=10.
Hope it helps:)

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3

Ah yes thank you so much!

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