CHEMISTRY HIGH SCHOOL

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. This is the equation: CO(g)+H2O(g)--><-- CO2(g) + H2(g). 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?

Answers

Answer 1

Answer:

Answer : The moles of $CO_2$ added will be 1.12 mole.

Solution : Given,

Moles of $CO$ and $H_2O$ at equilibrium = 0.200 mol

Moles of $CO_2$ and $H_2$ at equilibrium = 0.600 mol

First we have to calculate the concentration of $CO,H_2O,CO_2\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CO=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }H_2O=(Moles)/(Volume)=(0.200mol)/(1L)=0.200M$

$\text{Concentration of }CO_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

$\text{Concentration of }H_2=(Moles)/(Volume)=(0.600mol)/(1L)=0.600M$

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$K_c=((0.600)* (0.600))/((0.200)* (0.200))$

$K_c=9$

Now we have to calculate the moles of $CO_2$ added.

Let the moles of $CO_2$ added is 'x'.

The given equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initially 0.200 0.200 0.600 0.600

Added moles 0 0 x 0

Change +0.1 +0.1 -0.1 -0.1

Final 0.3 0.3 (0.5+x) 0.5

The expression of $K_c$ will be,

$K_c=([H_2][CO_2])/([CO][H_2O])$

$9=((0.5)* (0.5+x))/((0.3)* (0.3))$

$x=1.12mol$

Therefore, the moles of $CO_2$ added will be 1.12 mole.

Answer 2

Answer:

$\boxed{0.{\text{3 mol}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$ are added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 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:

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

Here,

K is the equilibrium constant.

P and Q are the reactants.

R and S are the products.

The given reaction is as follows:

${\text{CO}}\left(g\right) + {{\text{H}}_2}{\text{O}}\left( g \right)\rightleftharpoons {\text{C}}{{\text{O}}_2}\left( g \right) + {{\text{H}}_2}\left( g \right)$

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

${\text{K = }}\frac{{\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]\left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{\text{CO}}} \right]\left[ {{{\text{H}}_2}{\text{O}}}\right]}}$ ......(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 0.600 mol/L for $\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]$ , 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}} \right]$ , 0.200 mol/L for $\left[ {{\text{CO}}}\right]$ and 0.200 mol/L for $\left[ {{{\text{H}}_2}{\text{O}}} \right]$ in equation (1).

$\begin{aligned}{\text{K }}&=\frac{{\left( {{\text{0}}{\text{.600 mol/L}}}\right)\left( {{\text{0}}{\text{.600 mol/L}}}\right)}}{{\left( {{\text{0}}{\text{.200 mol/L}}}\right)\left( {{\text{0}}{\text{.200 mol/L}}}\right)}}\n&= 9\n\end{aligned}$

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

Rearrange equation (1) to calculate .

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

Substitute 9 for K, 0.300 mol/L for $\left[{{\text{CO}}}\right]$ , 0.200 mol/L for $\left[{{{\text{H}}_2}{\text{O}}}\right]$ and 0.600 mol/L for $\left[ {{{\text{H}}_{\text{2}}}}\right]$ in equation (2).

$\begin{aligned}\left[ {{\text{C}}{{\text{O}}_{\text{2}}}} \right]&=\frac{{{\text{9}}\left( {{\text{0}}{\text{.300 mol/L}}} \right)\left( {{\text{0}}{\text{.200 mol/L}}} \right)}}{{{\text{0}}{\text{.600 mol/L}}}}\n&= 0.{\text{9 mol/L}}\n\end{aligned}$

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:

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

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: brainly.com/question/3467841

2. Complete equation for the dissociation of (aq): 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.

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