The gas phase decompoistion of dimethyl ether follows first order kinetics.
`CH_(3)-O-CH_(3)(g)rarrCH_(4)(g)+CO(g)`
The reaction is carried out in a constant volume container at `500^(@)C` and has a half life of `14.5 min`. Initially, only dimethyl ether is present at a pressure `0.40 atm`. What is the total pressure of the system after `12min`? (Assume ideal gas behaviour)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the rate constant (k) from the half-life
The half-life (t₁/₂) for a first-order reaction is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Given that the half-life is 14.5 minutes, we can rearrange this equation to find k:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{14.5 \text{ min}} \]
Calculating this gives:
\[ k \approx 0.0478 \text{ min}^{-1} \]

### Step 2: Use the first-order kinetics equation to find the change in pressure (x)
For a first-order reaction, the relationship between the initial pressure (P₀), the remaining pressure (P), and the rate constant (k) is given by:
\[ k = \frac{2.303}{t} \log\left(\frac{P_0}{P}\right) \]
Here, \( P_0 = 0.40 \text{ atm} \) and \( t = 12 \text{ min} \). The remaining pressure can be expressed as:
\[ P = P_0 - x \]
where \( x \) is the change in pressure due to the decomposition of dimethyl ether.

Substituting into the equation:
\[ 0.0478 = \frac{2.303}{12} \log\left(\frac{0.40}{0.40 - x}\right) \]

### Step 3: Solve for x
Rearranging the equation gives:
\[ \log\left(\frac{0.40}{0.40 - x}\right) = \frac{0.0478 \times 12}{2.303} \]
Calculating the right side:
\[ \log\left(\frac{0.40}{0.40 - x}\right) \approx 0.249 \]

Now, we can exponentiate both sides to eliminate the logarithm:
\[ \frac{0.40}{0.40 - x} = 10^{0.249} \approx 1.77 \]
Cross-multiplying gives:
\[ 0.40 = 1.77(0.40 - x) \]
Expanding and rearranging:
\[ 0.40 = 0.708 - 1.77x \]
\[ 1.77x = 0.708 - 0.40 \]
\[ 1.77x = 0.308 \]
\[ x \approx \frac{0.308}{1.77} \approx 0.1746 \text{ atm} \]

### Step 4: Calculate the total pressure after 12 minutes
The total pressure in the system after 12 minutes can be calculated as:
\[ P_{total} = P_0 - x + 3x \]
Where \( 3x \) accounts for the production of methane and carbon monoxide (1 mole of dimethyl ether produces 1 mole of CH₄ and 1 mole of CO).

Substituting the values:
\[ P_{total} = 0.40 - 0.1746 + 3(0.1746) \]
\[ P_{total} = 0.40 - 0.1746 + 0.5238 \]
\[ P_{total} = 0.40 + 0.3492 \]
\[ P_{total} \approx 0.7492 \text{ atm} \]

### Final Answer:
The total pressure of the system after 12 minutes is approximately **0.7492 atm**.

---