Reaction Rate and Activation Energy

[Soaring Crane]

1) Methanol can be produced by the following reaction:
CO (g) + 2H2 (g) --> CH30H (g)

How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol?
-delta

/delta t = ?

a. + delta

/delta t
b. + 1delta[CH3OH]/2delta t
c. + 2 delta [CH3OH]/delta t
d. none of the above

Is it d because the rates are supposed to be determined experimentally?


2) The first order isomerization of cyclopropane -> propene has a rate constant of 0.00026 1/sec at 435 Celsius with an activation energy of 264 kJ/mol. What is the temperature at which the rate constant is 0.00447 1/sec? Calculate your answer in kelvins.

k2 = 0.00447 1/s
k1 = 0.00026 1/s
T1 = 435 C = 708.15 K
E_a = 264 kJ/mol = 2640 J/mol

ln (k2/k1) = (-E_a/R)*[(1-T2) - (1/T1)]?

1/T2 = [-ln (k2/k1)/(-E_a/R)] + (1/T1)
1/T2 = -ln(0.00447/.00026)/[-2640 J/mol/(8.314)] + (1/708.15 K)
=0.0096133567 + (1/708.15 K) = 0.011025

1/0.011025 = 90.69 K = 90.7 K?

Is the correct final answer and method?

Thank you.

 

[andrewchang]

number 1:
no, it has to do with the mole ratios.

number 2:
the equation should be:
$$ln(\frac{K_1}{K_2}) = \frac {E_a}{R} (\frac{1}{T_2}- \frac{1}{T_1})$$

 

[Soaring Crane]

1. So the rates are determined stoichiometrically? (I thought this was wrong.) The answer then would be b.??

2. ln (k2/k1) = (-E_a/R)*[(1-T2) - (1/T1)] I got this equation in my text. Why is the ratio of k's switched? (The - on the E_a/R will cancel out in solving for T2.)


Thanks.

 

[andrewchang]

1. no, hydrogen gas disappears at twice the rate

2. I'm sorry, I didn't notice the negative sign that you put in front of $E_a$

 

[GCT]

I'm pretty sure that

$$- \frac{d[H_2]}{2dt}= \frac{d[CH3OH]}{dt}$$

opposite signs just in case the latex isn't showing up