Question

An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

Answer 1

First you'll have to solve for the reaction equilibrium contrast, k. I believe that k=[Products]/[reactants], or 0.6^2/0.2^2 = 9
Now solve for the new concentration of CO2, X. 9=0.6*X/(0.3*0.2). This gives a CO2 concentration of 0.9 mol/L, meaning that 0.3 moles of CO2 have been added

Answer 2

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$  are added so as to increase the amount of carbon monoxide to 0.3 mol.

Further Explanation:

Chemical equilibrium is the state in which the concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{P(g)}} + {\text{Q(g)}}\rightleftharpoons {\text{R(g)}} + {\text{S(g)}}$

Equilibrium constant is the constant that relates to the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for the general reaction is as follows:

${\text{K}}=\dfrac{{\left[ {\text{R}} \right]\left[ {\text{S}}\right]}}{{\left[{\text{P}}\right]\left[ {\text{Q}} \right]}}$

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}} + {{\text{H}}_2}{\text{O}}\rightleftharpoons {\text{C}}{{\text{O}}_2} + {{\text{H}}_2}$

The expression for the equilibrium constant for the given reaction is as follows:

${\text{K = }}\dfrac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[{{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}}\right]\left[{{{\text{H}}_2}{\text{O}}} \right]}}$    .......(1)

Here,

K is the equilibrium constant.

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}}\right]$  is the concentration of carbon dioxide.

$\left[{{{\text{H}}_{\text{2}}}}\right]$ is the concentration of hydrogen.

$\left[{{\text{CO}}}\right]$ is the concentration of carbon monoxide.

$\left[{{{\text{H}}_2}{\text{O}}}\right]$ is the concentration of water.

Substitute 0.6 mol/L for $\left[{{\text{C}}{{\text{O}}_{\text{2}}}}\right]$ , 0.6 mol/L for $\left[{{{\text{H}}_{\text{2}}}}\right]$ , 0.2 mol/L for $\left[{{\text{CO}}}\right]$ and 0.2 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ in equation (1).

$\begin{aligned}{\text{K}}&=\frac{{\left( {{\text{0}}{\text{.6 mol/L}}} \right)\left( {{\text{0}}{\text{.6 mol/L}}} \right)}}{{\left({{\text{0}}{\text{.2 mol/L}}} \right)\left({{\text{0}}{\text{.2 mol/L}}} \right)}}\\&= 9\\\end{aligned}$

The value of equilibrium constant comes out to be 9 and it remains the same for the given reaction.

Rearrange equation (1) to calculate  .

$\left[{{\text{C}}{{\text{O}}_{\text{2}}}}\right]=\dfrac{{{\text{K}}\left( {\left[{{\text{CO}}} \right]\left[{{{\text{H}}_2}{\text{O}}}\right]}\right)}}{{\left[{{{\text{H}}_{\text{2}}}} \right]}}$            ......(2)

Substitute 9 for K, 0.3 mol/L for $\left[{{\text{CO}}}\right]$ , 0.2 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.6 mol/L for $\left[{{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}}\right]&= \frac{{{\text{9}}\left( {{\text{0}}{\text{.3 mol/L}}}\right)\left({{\text{0}}{\text{.2 mol/L}}}\right)}}{{{\text{0}}{\text{.6 mol/L}}}}\\&= 0.{\text{9 mol/L}}\\\end{aligned}$

Initially, 0.6 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ were present in a 1-L container. But now 0.9 moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ are present in it. So the extra amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ can be calculated as follows:

$\begin{aligned}{\text{Amount of C}}{{\text{O}}_{\text{2}}}{\text{ added}}&= 0.{\text{9 mol}} - {\text{6 mol}}\\&= 0.{\text{3 mol}}\\\end{aligned}$

Therefore 0.3 moles of carbon dioxide are added in a 1-L container.

Learn more:

1. Calculation of equilibrium constant of pure water at 25°c: https://brainly.com/question/3467841

2. Complete equation for the dissociation of  (aq): https://brainly.com/question/5425813

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Equilibrium

Keywords: CO, H2, CO2, H2O, 0.9 mol/L, 0.2 mol/L, 0.6 mol/L, 0.3 mol/L, K, carbon dioxide, water, hydrogen, carbon monoxide