Samples of $\ce{A}$ ($\pu{1.40 mol}$) and $\ce{B}$ ($\pu{0.80 mol}$) are placed in a $\pu{2.0 L}$ container and the following reaction takes place:

$$\ce{2 A (g) + B (g) <=> C (g) + D (g)}$$

At equilibrium, the concentration of $\ce{A}$ is $\pu{0.40 M}$. What is the value of $K_c$?

I thought you had to convert Sample B into molar concentrations using its mols and the 2.0 L volume, but i still do not know anything about C or D, and I can't use an ICE table because I am not given a Kc value. Do I assume molar concentrations are all the same because it is in equilibrium? I also tried using mole ratios to determine moles of C and D but I wasn't sure which mole value to use (A or B) and in the end it didn't work.

I have this question in a past paper sample exam. I know the answer already, but I require the solution as to know how to solve future questions similar to this.


closed as off-topic by Todd Minehardt, andselisk, Jon Custer, aventurin, Mithoron Feb 6 at 17:43

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  • $\begingroup$ You started with 1.40 moles of A and at equilibrium you have 0.8 moles of A. How many moles of A reacted? Then how many moles of B reacted? Finally, how many moles of C and D would you get from that reaction? @q123 $\endgroup$ – Tyberius Feb 6 at 15:24

First, start from a few basic definitions applicable in the case of this specific problem:

$$\ce{2 A (g) + B (g) <=> C (g) + DA (g)}$$

a) Relationship between number of moles, molar concentration, and volume

$$ [i] = \frac{n_i}{V} $$

b) Equilibrium constant in concentration units (when activity coefficients are ~1) and in terms of moles of each substance at equilibrium: $$ K_c = \frac{[C][DA]}{[A]^2[B]} = V\frac{n_{C,eq}n_{DA,eq}}{n_{A,eq}^2n_{B,eq}}$$


i) Compute final (equilibrium) moles of A, and moles of A that reacted:

$$ n_{A,eq} = [A]_{eq}V = 0.40 M \times 2.0 L =0.80 mole$$ $$ n_{A,reacted} = n_{A,init} -n_{A,eq} =(1.40 - 0.80) mole = 0.60 mole$$

ii) Compute moles of B that reacted and final moles of B:

$$ n_{B,reacted} = n_{A,reacted}/2 = 0.30 mole$$ $$ n_{B,eq} = n_{B,init}-n_{B,reacted} = (0.80 - 0.30) mole = 0.50 mole$$

iii) Compute moles of C and D that formed:

$$ n_{C/D,eq} = n_{C/D,formed} = n_{B,reacted}= 0.50 mole$$

iv) Put it all together into the expression for $K_c$

$$ K_c = 2.0 \frac{0.50^2}{0.80^20.50}M=1.56 M$$

Note in another common problem one sets up an equation starting from the definition of $K_c$, something like this:

$$ K_c = V \frac{x^2}{(n_A-2x)^2x}$$

and solves for an extent (or moles) of reaction x. In the case of this problem $K_c$ is unknown but x can be determined.


$$[\ce{A}]_I =\frac{1.4}{2}=\pu{0.7M},[\ce{B}]_I =\frac{0.8}{2}=\color{red}{\pu{0.4M}}$$

$$[\ce{A}]_e =\pu{0.4M}$$

$$[\ce{B}]_e =\color{red}{\pu{0.4}} -(\frac{(0.7-0.4)}{2})=\pu{0.25 M}=[\ce{C}]_e =[\ce{DA}]_e$$

$$ K_c = \frac{[C]_e[DA]_e}{[A]_e^2[B]_e} =\frac{[0.25][0.25]}{[0.4]^2[0.25]}=1.56 $$


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