The decomposition of arsine `(AsH_(3))` into arsenic and hudrogen is a first order reaction . The decomposition was studied at constant volume and at constant temperature. The pressures at different times are as follows :
`{:(t(h),0,5.5,6.5,8),(p(atm):,0.9654,1.06,1.076,1.1):}`
Calculate the velocity constant .

Explanation

To solve the problem of calculating the velocity constant (rate constant) for the decomposition of arsine (AsH₃) into arsenic and hydrogen, we will follow these steps:

### Step 1: Write the balanced reaction
The decomposition of arsine can be represented as:
\[ \text{2 AsH}_3 \rightarrow \text{2 As} + 3 \text{H}_2 \]

### Step 2: Understand the relationship between pressure and concentration
In a first-order reaction, the rate of reaction can be expressed in terms of the change in pressure. The total pressure at any time \( t \) can be expressed as:
\[ P_t = P_0 + \frac{3}{2}p \]
where:
- \( P_t \) is the total pressure at time \( t \)
- \( P_0 \) is the initial pressure
- \( p \) is the change in pressure due to the decomposition of arsine.

### Step 3: Gather initial and observed pressures
From the data given:
- At \( t = 0 \): \( P_0 = 0.9654 \, \text{atm} \)
- At \( t = 5.5 \, \text{h} \): \( P_t = 1.06 \, \text{atm} \)
- At \( t = 6.5 \, \text{h} \): \( P_t = 1.076 \, \text{atm} \)
- At \( t = 8 \, \text{h} \): \( P_t = 1.1 \, \text{atm} \)

### Step 4: Calculate the change in pressure \( p \)
Using the formula for total pressure:
\[ P_t = P_0 + \frac{3}{2}p \]
We can rearrange it to find \( p \):
\[ p = \frac{2}{3}(P_t - P_0) \]

Now, let's calculate \( p \) at \( t = 5.5 \, \text{h} \):
\[ p = \frac{2}{3}(1.06 - 0.9654) \]
\[ p = \frac{2}{3}(0.0946) \]
\[ p \approx 0.063 \, \text{atm} \]

### Step 5: Calculate the rate constant \( k \)
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{P_0 - p}{P_0}\right) \]

Substituting the values:
- \( P_0 = 0.9654 \, \text{atm} \)
- \( p = 0.063 \, \text{atm} \)
- \( t = 5.5 \, \text{h} \)

First, calculate \( P_0 - p \):
\[ P_0 - p = 0.9654 - 0.063 = 0.9024 \]

Now, substitute into the formula:
\[ k = \frac{2.303}{5.5} \log\left(\frac{0.9024}{0.9654}\right) \]

Calculating the logarithm:
\[ \log\left(\frac{0.9024}{0.9654}\right) \approx \log(0.934) \approx -0.031 \]

Now substitute back into the equation for \( k \):
\[ k = \frac{2.303}{5.5} \times (-0.031) \]
\[ k \approx -0.013 \, \text{h}^{-1} \]

### Final Result
The velocity constant \( k \) for the decomposition of arsine is approximately:
\[ k \approx 0.013 \, \text{h}^{-1} \]

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