Please show your work and answer for the following. Best answer gets the ten points immediately.
1. The gas-phase decomposition of SO2Cl2 = SO2 + Cl2 is first order in SO2Cl2. At 600 K, the half-life for this process is 2.3 x (10 to the fifth power) s. What is the rate constant at this temperature?
2. The fiirst order rate constant for the decomposition of N2O5, N2O5 = 2 NO2 + 1/2 O2, at 70 degrees celsius is 6.82 x (10 to the negative third power) s -1. Suppose we start with .03 moles of N2O5 in a volume of 2.5 L
a) How many moles of N2O5 will remain after 2.5 minutes?
b) How many minutes will it take forr this quantity of N2O5 to drop to .005 moles?
c) What is the half-life of N2O5 at 70 degrees celsius?
1. The gas-phase decomposition of SO2Cl2 = SO2 + Cl2 is first order in SO2Cl2. At 600 K, the half-life for this process is 2.3 x (10 to the fifth power) s. What is the rate constant at this temperature?
2. The fiirst order rate constant for the decomposition of N2O5, N2O5 = 2 NO2 + 1/2 O2, at 70 degrees celsius is 6.82 x (10 to the negative third power) s -1. Suppose we start with .03 moles of N2O5 in a volume of 2.5 L
a) How many moles of N2O5 will remain after 2.5 minutes?
b) How many minutes will it take forr this quantity of N2O5 to drop to .005 moles?
c) What is the half-life of N2O5 at 70 degrees celsius?
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this is the general form of a rate equation for A ---> B + C
rate = -d[A] / dt = k x [A]^n
if the reaction is 1st order.. n=1
-d[A] / dt = k x [A]
rearranging...
1/[A] d[A] = -kdt
integrating from initial concentration = [Ao] at time = 0 to final concentration = [At] at time = t
ln([At]) - ln([Ao]) = -kt
simplifying a bit...
ln([At] / [Ao]) = -kt
half live means time for 1/2 of "A" to be consumed.. ie [At] = 1/2 [Ao]...so...
ln(1/2[Ao] / [Ao]) = -kt
ln(1/2) = -kt
ln(1) - ln(2) = -kt
0 - ln(2) = -kt
k = ln(2) / t = ln(2) / (2.3x10^5 sec) = 3.01x10^-6 / s
****************
again.. 1st order means n=1 and
ln([At] / [Ao]) = -kt
a)..
ln([At] / [Ao]) = -kt
[At] / [Ao] = exp(-kt)
[At] = [Ao] x exp(-kt)
[At] = (0.03 moles / 2.5L) x exp(-6.82x10^-3/s x (2.5min x 60s/min))
[At] = 0.00431 moles / L
moles N2O5 = 2.5 L x (0.00431 moles / L) = 0.0108
>> notice that I divided by volume to get molarity and then multiplied to get moles? I could have skipped that right?
B).. same equation
ln([At] / [Ao]) = -kt
t = - ln(0.005 moles / 0.03 moles) / (6.82x10^-3 /s)
t = 262 seconds
C).. same equation... [At] = 1/2[Ao]
ln([At] / [Ao]) = -kt
ln(1/2 [Ao] / [Ao]) = -kt
ln(1/2) = -kt
ln(1) - ln(2) = -kt
ln(2) = kt
t = ln(2) / k = ln(2) / (6.82x10^-3 /s) = 102 seconds
****************
questions?
how do you suppose this would be different for a zero order reaction?
rate = -d[A] / dt = k x [A]^0
and for a 2nd order rxn
rate = -d[A] / dt = k x [A]^2
???
rate = -d[A] / dt = k x [A]^n
if the reaction is 1st order.. n=1
-d[A] / dt = k x [A]
rearranging...
1/[A] d[A] = -kdt
integrating from initial concentration = [Ao] at time = 0 to final concentration = [At] at time = t
ln([At]) - ln([Ao]) = -kt
simplifying a bit...
ln([At] / [Ao]) = -kt
half live means time for 1/2 of "A" to be consumed.. ie [At] = 1/2 [Ao]...so...
ln(1/2[Ao] / [Ao]) = -kt
ln(1/2) = -kt
ln(1) - ln(2) = -kt
0 - ln(2) = -kt
k = ln(2) / t = ln(2) / (2.3x10^5 sec) = 3.01x10^-6 / s
****************
again.. 1st order means n=1 and
ln([At] / [Ao]) = -kt
a)..
ln([At] / [Ao]) = -kt
[At] / [Ao] = exp(-kt)
[At] = [Ao] x exp(-kt)
[At] = (0.03 moles / 2.5L) x exp(-6.82x10^-3/s x (2.5min x 60s/min))
[At] = 0.00431 moles / L
moles N2O5 = 2.5 L x (0.00431 moles / L) = 0.0108
>> notice that I divided by volume to get molarity and then multiplied to get moles? I could have skipped that right?
B).. same equation
ln([At] / [Ao]) = -kt
t = - ln(0.005 moles / 0.03 moles) / (6.82x10^-3 /s)
t = 262 seconds
C).. same equation... [At] = 1/2[Ao]
ln([At] / [Ao]) = -kt
ln(1/2 [Ao] / [Ao]) = -kt
ln(1/2) = -kt
ln(1) - ln(2) = -kt
ln(2) = kt
t = ln(2) / k = ln(2) / (6.82x10^-3 /s) = 102 seconds
****************
questions?
how do you suppose this would be different for a zero order reaction?
rate = -d[A] / dt = k x [A]^0
and for a 2nd order rxn
rate = -d[A] / dt = k x [A]^2
???
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that's the math Rosalie. Do you have questions about it?
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