Suppose that the preceding reaction is carried out at a different temperature and that the initial concentrations of the reactants are $$ [\mathrm{CO}]=0.500 \mathrm{M} $$ and $$ \left[\mathrm{H}_{2}\right]=1.00 \mathrm{M} $$. Assuming there is no product at the beginning of the reaction, at equilibrium find the equilibrium $$ [\mathrm{CO}]=0.15 \mathrm{M} $$ constant at this new temperature. (Hint: Use the stoichiometric relationships from the balanced equation to find the equilibrium concentrations of $$ \mathrm{H}_{2} $$ and $$ \mathrm{CH}_{3} \mathrm{OH} $$.) Express your answer numerically.

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To find the equilibrium concentrations of $$ \mathrm{H}_2 $$ and $$ \mathrm{CH}_3\mathrm{OH} $$ given the equilibrium concentration of $$ \mathrm{CO} $$, we start with the balanced chemical equation for the reaction:

$$
\mathrm{CO} + 2\mathrm{H}_2 \rightleftharpoons \mathrm{CH}_3\mathrm{OH}
$$

### Initial Concentrations:
- $$ [\mathrm{CO}]_0 = 0.500 \, \mathrm{M} $$
- $$ [\mathrm{H}_2]_0 = 1.00 \, \mathrm{M} $$
- $$ [\mathrm{CH}_3\mathrm{OH}]_0 = 0 \, \mathrm{M} $$

### Change in Concentrations:
Let $$ x $$ be the change in concentration of $$ \mathrm{CO} $$ that reacts to reach equilibrium. According to the stoichiometry of the reaction:
- For every 1 mole of $$ \mathrm{CO} $$ that reacts, 2 moles of $$ \mathrm{H}_2 $$ react and 1 mole of $$ \mathrm{CH}_3\mathrm{OH} $$ is produced.

### Equilibrium Concentrations:
At equilibrium, we have:
- $$ [\mathrm{CO}] = [\mathrm{CO}]_0 - x = 0.500 - x $$
- $$ [\mathrm{H}_2] = [\mathrm{H}_2]_0 - 2x = 1.00 - 2x $$
- $$ [\mathrm{CH}_3\mathrm{OH}] = 0 + x = x $$

### Given Equilibrium Concentration:
We are given that at equilibrium, $$ [\mathrm{CO}] = 0.15 \, \mathrm{M} $$. Therefore, we can set up the equation:

$$
0.500 - x = 0.15
$$

Solving for $$ x $$:

$$
x = 0.500 - 0.15 = 0.350
$$

### Now, we can find the equilibrium concentrations of $$ \mathrm{H}_2 $$ and $$ \mathrm{CH}_3\mathrm{OH} $$:

1. **Equilibrium concentration of $$ \mathrm{H}_2 $$**:
   $$
   [\mathrm{H}_2] = 1.00 - 2x = 1.00 - 2(0.350) = 1.00 - 0.700 = 0.300 \, \mathrm{M}
   $$

2. **Equilibrium concentration of $$ \mathrm{CH}_3\mathrm{OH} $$**:
   $$
   [\mathrm{CH}_3\mathrm{OH}] = x = 0.350 \, \mathrm{M}
   $$

### Summary of Equilibrium Concentrations:
- $$ [\mathrm{CO}] = 0.15 \, \mathrm{M} $$
- $$ [\mathrm{H}_2] = 0.300 \, \mathrm{M} $$
- $$ [\mathrm{CH}_3\mathrm{OH}] = 0.350 \, \mathrm{M} $$

### Equilibrium Constant $$ K_c $$:
The equilibrium constant $$ K_c $$ for the reaction is given by:

$$
K_c = \frac{[\mathrm{CH}_3\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]^2}
$$

Substituting the equilibrium concentrations:

$$
K_c = \frac{0.350}{(0.15)(0.300)^2}
$$

Calculating $$ K_c $$:

$$
K_c = \frac{0.350}{(0.15)(0.09)} = \frac{0.350}{0.0135} \approx 25.93
$$

Thus, the equilibrium constant $$ K_c $$ at this new temperature is approximately **25.93**.
The equilibrium constant $$ K_c $$ at the new temperature is approximately 25.93.

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