Chemistry-Rate Law Problem?
The rate of the reaction CO2(g) + H2(g) → H2O(g) + CO(g) was determined in three experiments at 650 K. The results are given in the following table:
Experiment 1- [CO2] in mM=20.0, [H2] in mM=12.0, rate in mM/min= 2.3
Experiment 2- [CO2] in mM=20.0, [H2] in mM=24.0, rate in mM/min= 4.6
Experiment 3- [CO2] in mM=40.0, [H2] in mM=12.0, rate in mM/min= 4.6
A) What is the rate law for this reaction?
B) What is the rate constant k?
C) What would be the rate if starting concentrations were [CO2]=40.0 mM and [H2]=24.0 mM
- hcbiochemLv 710 months agoFavorite Answer
A) by comparing experiments 1 and 2, you see that doubling [H2] doubles the rate. So, the reaction is first order with respect to [H2]. By comparing reactions 1 and 3, you see that doubling [CO2] also doubles the rate, so the reaction is also first order with respect to [CO2]. Therefore, the rate law is:
Rate = k[CO2][H2]
B) To calculate the value of k, just plug in everything from one of the experiments into the rate law and solve for k:
2.3 mM/min = k(20.0)(12.0)
k = 9.6X10^-3 mM^-1 min^-1
C) Again, just plug what you know into the rate law and calculate the rate:
Rate = 9.6X10^-3 mM^-1 min^-1 (40.0 mM) (24.0 mM)
Rate = 9.2 mM/min