the decomposition of (CH3)2O has a rate constant of 6.8x10^-4 s^-1. the initial pressure is 135 torr, what is the partial pressure after 1420s?
i don't have the slightest idea on how to solve this.
To solve this question, you need to apply the first-order rate equation to calculate the partial pressure after a given time. Let's break down the steps:
Step 1: Recall the first-order rate equation:
ln(Pt/P0) = -kt
Where:
Pt = final pressure
P0 = initial pressure
k = rate constant
t = time
Step 2: Rearrange the equation to solve for Pt:
Pt = P0 * e^(-kt)
Step 3: Plug in the given values:
P0 = 135 torr
k = 6.8 × 10^-4 s^-1
t = 1420s
Step 4: Calculate Pt using the equation above:
Pt = 135 torr * e^(-6.8 × 10^-4 s^-1 * 1420s)
Now, let's calculate the partial pressure after 1420s:
1. Convert the rate constant to minutes^-1 (since time is given in seconds):
k = 6.8 × 10^-4 s^-1 * 60s/min
= 4.08 × 10^-2 min^-1
2. Plug in the given values and calculate Pt using the equation:
Pt = 135 torr * e^(-4.08 × 10^-2 min^-1 * 1420s)
Using a scientific calculator, perform the exponential calculation, and you'll find the partial pressure after 1420s.
To solve this problem, we will use the first-order reaction decay equation:
\[ P = P_0 \times e^{-kt} \]
Where:
- P is the partial pressure at time t
- P0 is the initial partial pressure
- k is the rate constant
- t is the time
Let's plug in the given values:
\[ P = 135 \, \text{torr} \times e^{-(6.8 \times 10^{-4} \, \text{s}^{-1}) \times 1420 \, \text{s}} \]
Let's calculate this step-by-step:
Step 1: Calculate the exponent using the rate constant and time:
\[ -(6.8 \times 10^{-4} \, \text{s}^{-1}) \times 1420 \, \text{s} = -0.9656 \]
Step 2: Calculate the value of Euler's number, e, raised to the exponent computed in Step 1 using a calculator:
\[ e^{-0.9656} \approx 0.3817 \]
Step 3: Multiply the initial partial pressure by the value obtained in Step 2 to get the partial pressure at the given time:
\[ P = 135 \, \text{torr} \times 0.3817 \]
Step 4: Calculate the partial pressure:
\[ P \approx 51.52 \, \text{torr} \]
Therefore, after 1420 seconds, the partial pressure is approximately 51.52 torr.