# Molarity and equilibrium reactions Kc [closed]

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.

• 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 – Tyberius Feb 6 at 15:24

I thought you had to convert Sample B into molar concentrations using its mols and the 2.0 L volume,

Yes,you can calculate the concentration of $$B$$ at equilibrium by the following expression: $$[\ce{B}]_\mathrm{e}= \frac{(n_{(B)})_e}{V}$$

Do I assume molar concentrations are all the same because it is in equilibrium?

NO ,you must calculate the molar concentrations of each species at equilibrium.

but i still do not know anything about C or D,

• The initial amount of C =The initial amount of D =zero

• The amount of C at equilibrium =The amount of C formed from the reaction

• The amount of D at equilibrium =The amount of D formed from the reaction

I also tried using mole ratios to determine moles of C and D

\begin{align}\text{ The amount of C at equilibrium} &=\text{The amount of D at equilibrium} \\&= \text{The amount of B reacted} \\ &= \frac{\text{ The amount of A reacted}}{2} \end{align}

but I wasn't sure which mole value to use (A or B) and in the end it didn't work.

• You have the initial amount of A equal $$\pu{1.40 mol}$$ ,the initial amount of B equal $$\pu{0.8 mol}$$ and the concentration of A at equilibrium equal $$\pu{0.40 M}=\pu{0.40 \frac{mol}{L}}$$.

• Calculate the amount of A at equilibrium :

\begin{align}\text{Amount of A at equilibrium} &=\text{concentration of A at equilibrium}\times{\text{Volume of the container}}\\ (n_\mathrm{A})_e &= C_\mathrm{A}\times{V}\\ &=\pu{0.40\frac{mol}{L}}\times{\pu{2L}}\\&=\pu{0.8mol} \end{align}

• Calculate the amount of A reacted : \begin{align}\text{Amount of A reacted}&=\text{Initial amount of A }-\text{Amount of A at equilibrium}\\ (n_{(A)})_\text{reacted} &=(n_{(A)})_I -(n_{(A)})_e\\&=1.4-0.8\\&=\pu{0.6mol}\end{align}

• Calculate the amount of B reacted : \begin{align} \text{The amount of B reacted} &= \frac{\text{ The amount of A reacted}}{2}\\ &= \frac{\pu{0.6mol}}{2}\\&=\pu{0.3mol}\\&=\text{The amount of C at equilibrium}\\ &=\text{The amount of D at equilibrium} \end{align}

• Calculate the amount of B at equilibrium : \begin{align}\text{Amount of B at equilibrium}&=\text{Initial amount of B }-\text{Amount of B reacted}\\ (n_{(B)})_e &=(n_{(B)})_I -(n_{(B)})_\text{reacted}\\&=0.8-0.3\\&=\pu{0.5mol} \end{align}

and I can't use an ICE table because I am not given a Kc value.

• You can use the following table to determine the amount of each species at equilibrium in order to calculate $$K_\mathrm{C}$$ : \begin{align} &&\ce{2A_{(g)} &&+ &&B_{(g)}&<=>&C_{(g)} &+&&D_{(g)}\\ I &&(1.4\pu{mol}) &&&&(0.8\pu{mol}) &&(0\pu{mol}) &&&(0\pu{mol})\\ C &&(-0.6\pu{mol}) &&&&(-0.3\pu{mol}) &&(0.3\pu{mol}) &&&(0.3\pu{mol})\\ E &&(0.8\pu{mol}) &&&&(0.5\pu{mol}) &&(0.3\pu{mol}) &&&(0.3\pu{mol})} \end{align}

• Use the expression $$[i]_\mathrm{e}= \frac{(n_{(i)})_e}{V}$$ to calculate the concentration of each species at equilibrium: \begin{align}[\mathrm{A}]_\mathrm{e}= \frac{\pu{0.8mol}}{\pu{2L}}=\pu{0.4M}\\ [\mathrm{B}]_\mathrm{e}= \frac{\pu{0.5mol}}{\pu{2L}}=\pu{0.25M}\\ [\mathrm{C}]_\mathrm{e}= \frac{\pu{0.3mol}}{\pu{2L}}=\pu{0.15M}\\ [\mathrm{D}]_\mathrm{e}= \frac{\pu{0.3mol}}{\pu{2L}}=\pu{0.15M}\\ \end{align}
• Substitute the concentration of each species at equilibrium in the following formula to calculate $$K_\mathrm{C}$$: \begin{align} K_\mathrm{C}&=\frac{[\mathrm{C}]_\mathrm{e}[\mathrm{D}]_\mathrm{e} }{[\mathrm{A}]^2_\mathrm{e}[\mathrm{B}]_\mathrm{e} }\\ &=\frac{(\pu{0.15M})(\pu{0.15M}) }{(\pu{0.4M)^2}(\pu{0.25M}) }\\ &=0.5625\pu{\frac{L}{mol}} \end{align}

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: $$K_c = \frac{[C][DA]}{[A]^2[B]} = V\frac{n_{C,eq}n_{DA,eq}}{n_{A,eq}^2n_{B,eq}}$$

Strategy:

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:

$$n_{C/D,eq} = n_{C/D,formed} = n_{B,reacted}= 0.30 mole$$

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

$$K_c = 2.0 \frac{0.30^2}{0.80^20.50}M=0.56 M$$

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.

• :"nC/D,eq=nC/D,formed=nB,reacted=0.50mole".I think nC/D,eq=0.30mol – Adnan AL-Amleh Mar 25 at 20:08