What is the rate constant when given pressure and time for a 1st order reaction?
How would I do this?
Here is the question:
The following gas-phase reaction was studied at 290°C by observing the change in pressure as a function of time in a constant-volume vessel.
ClCO2CCl3 (g) → 2COCl2 (g)
Time (s)___P (mmHg)
0________15.76
181______18.88
513______22.79
1164_____27.08
Where P is the total pressure
From the data, I was able to find out that it was a 1st order reaction when I graphed it. I'm just having trouble finding the rate constant (which should be 1.08x10^-3 s^-1)
How would I do this?
Here is the question:
The following gas-phase reaction was studied at 290°C by observing the change in pressure as a function of time in a constant-volume vessel.
ClCO2CCl3 (g) → 2COCl2 (g)
Time (s)___P (mmHg)
0________15.76
181______18.88
513______22.79
1164_____27.08
Where P is the total pressure
From the data, I was able to find out that it was a 1st order reaction when I graphed it. I'm just having trouble finding the rate constant (which should be 1.08x10^-3 s^-1)
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When reaction problems like this involve pressure they can be a little confusing.
The first thing you have to do is determine the partial pressure of the gas COCl2. Since 2 moles COCl2 are made for every one mole ClCO2CCl3, you can represent the partial pressure of COCl2 as:
P(COCl2) = P(o) Total - (P Total - P(o) Total)
Where P(o) Total is the original total pressure, and P Total is the total pressure at each given time.
Using that formula you get the following partial pressure values:
15.76 @ 0 sec
12.64 @ 181 sec
8.73 @ 573 sec
4.44 @ 1164 sec
You plot the natural log of these values against time and the slope will equal -k, so just take the negative sign off to find the rate constant of 1.08*10^-3 s^-1
Now you could also use the integrated first order rate law to solve for k without graphing:
ln Pt = -k(t) + ln Po
Where Pt is partial pressure at time t, t is time, and Po is initial partial pressure.
Rearrange: (ln(Pt/Po))/t = -k
ln (4.44/15.76) / 1164 = -k
k= 1.08*10^3 s^-1
Hope that helps you out.
The first thing you have to do is determine the partial pressure of the gas COCl2. Since 2 moles COCl2 are made for every one mole ClCO2CCl3, you can represent the partial pressure of COCl2 as:
P(COCl2) = P(o) Total - (P Total - P(o) Total)
Where P(o) Total is the original total pressure, and P Total is the total pressure at each given time.
Using that formula you get the following partial pressure values:
15.76 @ 0 sec
12.64 @ 181 sec
8.73 @ 573 sec
4.44 @ 1164 sec
You plot the natural log of these values against time and the slope will equal -k, so just take the negative sign off to find the rate constant of 1.08*10^-3 s^-1
Now you could also use the integrated first order rate law to solve for k without graphing:
ln Pt = -k(t) + ln Po
Where Pt is partial pressure at time t, t is time, and Po is initial partial pressure.
Rearrange: (ln(Pt/Po))/t = -k
ln (4.44/15.76) / 1164 = -k
k= 1.08*10^3 s^-1
Hope that helps you out.