`CH_(3)O CH_(3) (g) to CH_(4)(g) + H_(2)(g) + CO(g)` . The decomposition follows first order kietics at `500^(@)C` and has half - life 14.5 min. Initially only dimethyl ether is present at a pressure of 0.4 atm. Calculate the pressure of the mixture after 12 min.

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Understand the Reaction
The reaction is given as:
\[ \text{CH}_3\text{OCH}_3 (g) \rightarrow \text{CH}_4 (g) + \text{H}_2 (g) + \text{CO} (g) \]

### Step 2: Initial Conditions
- Initial pressure of dimethyl ether, \( P_0 = 0.4 \, \text{atm} \)
- At \( t = 0 \), the pressures of the products (CH₄, H₂, CO) are all \( 0 \, \text{atm} \).

### Step 3: Change in Pressure Over Time
Let \( x \) be the change in pressure of dimethyl ether that has decomposed after 12 minutes. Thus, at \( t = 12 \, \text{min} \):
- Pressure of dimethyl ether = \( 0.4 - x \)
- Pressure of CH₄ = \( x \)
- Pressure of H₂ = \( x \)
- Pressure of CO = \( x \)

### Step 4: Total Pressure After 12 Minutes
The total pressure \( P_t \) at \( t = 12 \, \text{min} \) can be expressed as:
\[ P_t = (0.4 - x) + x + x + x = 0.4 - x + 3x = 0.4 + 2x \]

### Step 5: Calculate the Rate Constant \( k \)
Given the half-life \( t_{1/2} = 14.5 \, \text{min} \) for a first-order reaction:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{14.5} \approx 0.0478 \, \text{min}^{-1} \]

### Step 6: Use the First-Order Kinetics Formula
For a first-order reaction, the rate constant \( k \) can also be expressed as:
\[ k = \frac{2.303}{t} \log\left(\frac{P_0}{P_t}\right) \]
Substituting the known values:
\[ 0.0478 = \frac{2.303}{12} \log\left(\frac{0.4}{0.4 - x}\right) \]

### Step 7: Solve for \( x \)
Rearranging and solving for \( x \):
1. Multiply both sides by 12:
\[ 0.5736 = 2.303 \log\left(\frac{0.4}{0.4 - x}\right) \]
2. Divide by 2.303:
\[ \log\left(\frac{0.4}{0.4 - x}\right) = 0.249 \]
3. Convert from logarithmic form:
\[ \frac{0.4}{0.4 - x} = 10^{0.249} \approx 1.778 \]
4. Cross-multiply to solve for \( x \):
\[ 0.4 = 1.778(0.4 - x) \]
\[ 0.4 = 0.7112 - 1.778x \]
\[ 1.778x = 0.7112 - 0.4 \]
\[ 1.778x = 0.3112 \]
\[ x \approx 0.1746 \, \text{atm} \]

### Step 8: Calculate Total Pressure
Substituting \( x \) back into the total pressure equation:
\[ P_t = 0.4 + 2(0.1746) \]
\[ P_t = 0.4 + 0.3492 = 0.7492 \, \text{atm} \]

### Final Answer
The pressure of the mixture after 12 minutes is approximately:
\[ P_t \approx 0.7492 \, \text{atm} \]

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