Answer :
Let's solve the given problem step-by-step for the decomposition of [tex]\( \text{N}_2\text{O}_5 \)[/tex] using the information provided:
### Step 1: Write down the given values
- Rate constant, [tex]\( k = 6.82 \times 10^{-3} \, \text{s}^{-1} \)[/tex]
- Initial amount of [tex]\( \text{N}_2\text{O}_5 \)[/tex], [tex]\( \text{initial\_mol} = 0.300 \, \text{mol} \)[/tex]
- Volume of the container, [tex]\( V = 0.500 \, \text{L} \)[/tex]
- Time, [tex]\( t = 1.5 \, \text{min} \)[/tex]
### Step 2: Convert the time from minutes to seconds
Since the rate constant [tex]\( k \)[/tex] is in units of [tex]\(\text{s}^{-1}\)[/tex], we need to convert the time to seconds:
[tex]\[ \text{time\_s} = t \times 60 \, \text{s/min} = 1.5 \, \text{min} \times 60 = 90 \, \text{s} \][/tex]
### Step 3: Calculate initial concentration
The initial concentration ([tex]\([A_0]\)[/tex]) of [tex]\( \text{N}_2\text{O}_5 \)[/tex] can be calculated using the formula:
[tex]\[ \text{concentration\_initial} = \frac{\text{initial\_mol}}{V} = \frac{0.300 \, \text{mol}}{0.500 \, \text{L}} = 0.600 \, \text{mol/L} \][/tex]
### Step 4: Use the first-order reaction formula
The formula for a first-order reaction is:
[tex]\[ [A] = [A_0] \cdot e^{-kt} \][/tex]
where:
- [tex]\([A_0]\)[/tex] is the initial concentration,
- [tex]\( k \)[/tex] is the rate constant, and
- [tex]\( t \)[/tex] is the time.
Plugging in the values we have:
[tex]\[ [A] = 0.600 \, \text{mol/L} \cdot e^{- (6.82 \times 10^{-3} \, \text{s}^{-1} \times 90 \, \text{s})} \][/tex]
### Step 5: Calculate the remaining concentration
[tex]\[ [A] \approx 0.600 \, \text{mol/L} \cdot e^{-0.6138} \][/tex]
Using a calculator or an exponentiation function:
[tex]\[ [A] \approx 0.3247740322805302 \, \text{mol/L} \][/tex]
### Step 6: Find the remaining moles
Now, we need to find how many moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] remain in the container after 1.5 minutes. This is given by:
[tex]\[ \text{remaining\_mol} = [A] \cdot V = 0.3247740322805302 \, \text{mol/L} \cdot 0.500 \, \text{L} = 0.1623870161402651 \, \text{mol} \][/tex]
### Answer
The number of moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] remaining after 1.5 minutes is approximately [tex]\( 0.1624 \, \text{mol} \)[/tex].
### Step 1: Write down the given values
- Rate constant, [tex]\( k = 6.82 \times 10^{-3} \, \text{s}^{-1} \)[/tex]
- Initial amount of [tex]\( \text{N}_2\text{O}_5 \)[/tex], [tex]\( \text{initial\_mol} = 0.300 \, \text{mol} \)[/tex]
- Volume of the container, [tex]\( V = 0.500 \, \text{L} \)[/tex]
- Time, [tex]\( t = 1.5 \, \text{min} \)[/tex]
### Step 2: Convert the time from minutes to seconds
Since the rate constant [tex]\( k \)[/tex] is in units of [tex]\(\text{s}^{-1}\)[/tex], we need to convert the time to seconds:
[tex]\[ \text{time\_s} = t \times 60 \, \text{s/min} = 1.5 \, \text{min} \times 60 = 90 \, \text{s} \][/tex]
### Step 3: Calculate initial concentration
The initial concentration ([tex]\([A_0]\)[/tex]) of [tex]\( \text{N}_2\text{O}_5 \)[/tex] can be calculated using the formula:
[tex]\[ \text{concentration\_initial} = \frac{\text{initial\_mol}}{V} = \frac{0.300 \, \text{mol}}{0.500 \, \text{L}} = 0.600 \, \text{mol/L} \][/tex]
### Step 4: Use the first-order reaction formula
The formula for a first-order reaction is:
[tex]\[ [A] = [A_0] \cdot e^{-kt} \][/tex]
where:
- [tex]\([A_0]\)[/tex] is the initial concentration,
- [tex]\( k \)[/tex] is the rate constant, and
- [tex]\( t \)[/tex] is the time.
Plugging in the values we have:
[tex]\[ [A] = 0.600 \, \text{mol/L} \cdot e^{- (6.82 \times 10^{-3} \, \text{s}^{-1} \times 90 \, \text{s})} \][/tex]
### Step 5: Calculate the remaining concentration
[tex]\[ [A] \approx 0.600 \, \text{mol/L} \cdot e^{-0.6138} \][/tex]
Using a calculator or an exponentiation function:
[tex]\[ [A] \approx 0.3247740322805302 \, \text{mol/L} \][/tex]
### Step 6: Find the remaining moles
Now, we need to find how many moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] remain in the container after 1.5 minutes. This is given by:
[tex]\[ \text{remaining\_mol} = [A] \cdot V = 0.3247740322805302 \, \text{mol/L} \cdot 0.500 \, \text{L} = 0.1623870161402651 \, \text{mol} \][/tex]
### Answer
The number of moles of [tex]\( \text{N}_2\text{O}_5 \)[/tex] remaining after 1.5 minutes is approximately [tex]\( 0.1624 \, \text{mol} \)[/tex].
