this is a text book question that i cannot figure out.
the decomposition of dimethyl ether at 510 degrees is a first-order process with a rate constant of 6.8*10^-4.
(CH3)2O(g)--> CH4(g) + H2(g) + CO(g)
if the partial pressure of (CH3)20 is 18kPa, what is its partial pressure after 1420sec.
I think you can simply substitute into
ln(Po/P) = kt and solve for p.
To find the partial pressure of (CH3)2O after 1420 seconds, we can use the first-order rate equation:
Rate = k * [A]
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of the reactant
In this case, since we are given the partial pressure instead of the concentration, we can use the partial pressure instead of concentration in the rate equation.
Given information:
- K = 6.8 * 10^-4 (rate constant)
- The reaction is a first-order process, which means the rate is directly proportional to the concentration (or partial pressure) of (CH3)2O.
We can rearrange the rate equation to solve for [A]:
Rate = k * [A]
[A] = Rate / k
Now, let's calculate the partial pressure of (CH3)2O after 1420 seconds.
Step 1: Calculate the rate of the reaction.
We are not given the rate directly, but we know that the reaction is first-order, meaning the rate is proportional to [A] (or the partial pressure of (CH3)2O). Therefore, we can assume that the rate at any given moment is directly proportional to the partial pressure of (CH3)2O.
We can express this relationship as:
Rate = k * (partial pressure of (CH3)2O)
Given:
k = 6.8 * 10^-4 (rate constant)
partial pressure of (CH3)2O = 18 kPa
Rate = (6.8 * 10^-4) * (18)
Rate = 0.01224 kPa/s
Step 2: Calculate the partial pressure after 1420 seconds.
partial pressure after 1420 seconds = partial pressure initially + (rate * time)
Given:
partial pressure initially = 18 kPa
time = 1420 seconds
rate = 0.01224 kPa/s
partial pressure after 1420 seconds = 18 + (0.01224 * 1420)
partial pressure after 1420 seconds ≈ 18 + 17.4048
partial pressure after 1420 seconds ≈ 35.4048 kPa
Therefore, the partial pressure of (CH3)2O after 1420 seconds is approximately 35.4048 kPa.