Answer :
To solve for the equilibrium concentration of [tex]\( CO_2(g) \)[/tex] in the reaction:
[tex]\[ CO(g) + H_2O(g) \leftrightharpoons CO_2(g) + H_2(g) \][/tex]
we will use the following steps:
1. Initial Concentrations:
- The initial concentration of [tex]\( CO(g) \)[/tex] is 0.400 M.
- The initial concentration of [tex]\( H_2O(g) \)[/tex] is 0.400 M.
- The initial concentrations of [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] are both 0 M.
2. Change in Concentration:
- Let [tex]\( x \)[/tex] be the change in concentration for [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] at equilibrium.
- Therefore, the concentration of [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] at equilibrium will be [tex]\( x \)[/tex].
3. Equilibrium Concentrations:
- At equilibrium, the concentration of [tex]\( CO(g) \)[/tex] will be [tex]\( 0.400 - x \)[/tex].
- The concentration of [tex]\( H_2O(g) \)[/tex] will also be [tex]\( 0.400 - x \)[/tex].
- The concentrations of both [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] will be [tex]\( x \)[/tex].
4. Equilibrium Expression:
The equilibrium constant [tex]\( K \)[/tex] for the reaction is given by:
[tex]\[
K = \frac{[CO_2][H_2]}{[CO][H_2O]}
\][/tex]
Substitute the equilibrium concentrations into the expression:
[tex]\[
98.2 = \frac{x \times x}{(0.400 - x)(0.400 - x)}
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
Rearrange and solve the equation for [tex]\( x \)[/tex].
6. Equilibrium Concentration of [tex]\( CO_2(g) \)[/tex]:
The equilibrium concentration of [tex]\( CO_2(g) \)[/tex] is [tex]\( x \)[/tex].
After solving the equation, we find that the equilibrium concentration of [tex]\( CO_2(g) \)[/tex] is approximately [tex]\( 0.363 \, M \)[/tex].
Thus, the best approach was to use the equilibrium expression and solve for [tex]\( x \)[/tex], giving us the equilibrium concentration of [tex]\( CO_2(g) \)[/tex] as [tex]\( 0.363 \, M \)[/tex].
[tex]\[ CO(g) + H_2O(g) \leftrightharpoons CO_2(g) + H_2(g) \][/tex]
we will use the following steps:
1. Initial Concentrations:
- The initial concentration of [tex]\( CO(g) \)[/tex] is 0.400 M.
- The initial concentration of [tex]\( H_2O(g) \)[/tex] is 0.400 M.
- The initial concentrations of [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] are both 0 M.
2. Change in Concentration:
- Let [tex]\( x \)[/tex] be the change in concentration for [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] at equilibrium.
- Therefore, the concentration of [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] at equilibrium will be [tex]\( x \)[/tex].
3. Equilibrium Concentrations:
- At equilibrium, the concentration of [tex]\( CO(g) \)[/tex] will be [tex]\( 0.400 - x \)[/tex].
- The concentration of [tex]\( H_2O(g) \)[/tex] will also be [tex]\( 0.400 - x \)[/tex].
- The concentrations of both [tex]\( CO_2(g) \)[/tex] and [tex]\( H_2(g) \)[/tex] will be [tex]\( x \)[/tex].
4. Equilibrium Expression:
The equilibrium constant [tex]\( K \)[/tex] for the reaction is given by:
[tex]\[
K = \frac{[CO_2][H_2]}{[CO][H_2O]}
\][/tex]
Substitute the equilibrium concentrations into the expression:
[tex]\[
98.2 = \frac{x \times x}{(0.400 - x)(0.400 - x)}
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
Rearrange and solve the equation for [tex]\( x \)[/tex].
6. Equilibrium Concentration of [tex]\( CO_2(g) \)[/tex]:
The equilibrium concentration of [tex]\( CO_2(g) \)[/tex] is [tex]\( x \)[/tex].
After solving the equation, we find that the equilibrium concentration of [tex]\( CO_2(g) \)[/tex] is approximately [tex]\( 0.363 \, M \)[/tex].
Thus, the best approach was to use the equilibrium expression and solve for [tex]\( x \)[/tex], giving us the equilibrium concentration of [tex]\( CO_2(g) \)[/tex] as [tex]\( 0.363 \, M \)[/tex].
