The gas phase decomposition of dimethylether follows first order kinetics `CH_3-O-CH_3(g)rarrCH_4(g)+H_2(g)+CO_((g))` The reaction is carried out in constant volume container at `500^@C` and has a half - life of 14.5 . Initially only dimethylether is present at a pressure of 0.40 atm . The total pressure of the system after 12 min is `x/100` atm . The value of x is [Given `10^(0.25)=1.778`]

Answer

To solve the problem step-by-step, we will follow the outlined procedure based on the information given in the question.

### Step 1: Understand the Reaction and Initial Conditions
The decomposition of dimethyl ether (\(CH_3OCH_3\)) is given by the reaction:
\[ CH_3OCH_3(g) \rightarrow CH_4(g) + H_2(g) + CO(g) \]

Initially, we have:
- Pressure of dimethyl ether, \(P_{initial} = 0.40 \, \text{atm}\)
- At \(t = 0\), the pressures of \(CH_4\), \(H_2\), and \(CO\) are all \(0 \, \text{atm}\).

### Step 2: Determine the Rate Constant \(k\)
The half-life (\(t_{1/2}\)) for a first-order reaction is given as \(14.5 \, \text{min}\). The formula for the half-life of a first-order reaction is:
\[ t_{1/2} = \frac{0.693}{k} \]

Rearranging this gives:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{14.5} \approx 0.0478 \, \text{min}^{-1} \]

### Step 3: Calculate the Change in Pressure After 12 Minutes
Using the first-order kinetics equation:
\[ k = \frac{2.303}{T} \log\left(\frac{P_{initial}}{P_{final}}\right) \]

Where \(T\) is the time in minutes. We need to find \(P_{final}\) after \(12 \, \text{min}\):
- Let \(x\) be the change in pressure of \(CH_3OCH_3\) that decomposes. Thus, after \(12 \, \text{min}\):
\[ P_{final} = P_{initial} - x = 0.40 - x \]

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

### Step 4: Solve for \(x\)
Rearranging gives:
\[ \log\left(\frac{0.40}{0.40 - x}\right) = 0.0478 \cdot \frac{12}{2.303} \]
Calculating the right side:
\[ 0.0478 \cdot \frac{12}{2.303} \approx 0.249 \]

Now, we have:
\[ \frac{0.40}{0.40 - x} = 10^{0.249} \]
Using the given value \(10^{0.25} \approx 1.778\):
\[ 10^{0.249} \approx 1.774 \]

Thus:
\[ \frac{0.40}{0.40 - x} = 1.774 \]
Cross-multiplying gives:
\[ 0.40 = 1.774(0.40 - x) \]
Expanding:
\[ 0.40 = 0.7096 - 1.774x \]
Rearranging to isolate \(x\):
\[ 1.774x = 0.7096 - 0.40 \]
\[ 1.774x = 0.3096 \]
\[ x = \frac{0.3096}{1.774} \approx 0.1745 \]

### Step 5: Calculate Total Pressure After 12 Minutes
The total pressure after \(12 \, \text{min}\) is given by:
\[ P_{total} = (0.40 - x) + x + x + x = 0.40 - x + 3x = 0.40 + 2x \]

Substituting \(x\):
\[ P_{total} = 0.40 + 2(0.1745) \]
\[ P_{total} = 0.40 + 0.349 = 0.749 \, \text{atm} \]

### Step 6: Relate Total Pressure to \(x\)
According to the problem, the total pressure after \(12 \, \text{min}\) is given as:
\[ P_{total} = \frac{x}{100} \]

Setting this equal to our calculated total pressure:
\[ \frac{x}{100} = 0.749 \]
Thus:
\[ x = 0.749 \times 100 = 74.9 \]

Rounding gives \(x = 75\).

### Final Answer
The value of \(x\) is **75**.