A first order gas reaction has `k = 1.5 xx 10^(-6) s^(-1)` at `200^(@)C`. If the reaction is allowed to run for `10 h`, what have changed in the Product? What is the half-life of this reaction?

Explanation

To solve the problem, we need to determine two things: the change in the product after the reaction runs for 10 hours and the half-life of the reaction.

### Step 1: Convert Time from Hours to Seconds
Given that the reaction runs for 10 hours, we first convert this time into seconds:
\[ 10 \text{ hours} = 10 \times 3600 \text{ seconds} = 36000 \text{ seconds} \]

**Hint:** Remember that 1 hour equals 3600 seconds.

### Step 2: Use the First Order Reaction Rate Equation
For a first-order reaction, the rate constant \( k \) is related to the concentrations of reactants and products by the equation:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
Where:
- \( k = 1.5 \times 10^{-6} \, \text{s}^{-1} \)
- \( t = 36000 \, \text{s} \)
- \( [A_0] \) is the initial concentration
- \( [A] \) is the remaining concentration after time \( t \)

Substituting the known values into the equation:
\[ 1.5 \times 10^{-6} = \frac{2.303}{36000} \log \left( \frac{[A_0]}{[A]} \right) \]

### Step 3: Solve for the Logarithmic Term
Rearranging the equation gives:
\[ \log \left( \frac{[A_0]}{[A]} \right) = \frac{1.5 \times 10^{-6} \times 36000}{2.303} \]
Calculating the right side:
\[ \log \left( \frac{[A_0]}{[A]} \right) \approx \frac{54}{2.303} \approx 23.5 \]

### Step 4: Exponentiate to Find the Concentration Ratio
Now we can find the ratio of initial to remaining concentration:
\[ \frac{[A_0]}{[A]} = 10^{23.5} \]

### Step 5: Calculate Change in Product
Let’s denote the change in concentration as \( x \):
\[ [A_0] - [A] = x \]
From the ratio, we can express:
\[ [A] = \frac{[A_0]}{10^{23.5}} \implies x = [A_0] - \frac{[A_0]}{10^{23.5}} = [A_0] \left(1 - \frac{1}{10^{23.5}}\right) \]
This indicates that a very small fraction of the reactant has reacted, leading to a change in product concentration.

### Step 6: Calculate the Half-Life of the Reaction
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{1.5 \times 10^{-6}} \approx 462000 \text{ seconds} \approx 128.33 \text{ hours} \]

### Summary of Results
1. The change in product concentration after 10 hours is approximately \( 4.7\% \) of the initial concentration.
2. The half-life of the reaction is approximately \( 128.33 \) hours.