A chemistry student wants to determine the rate constant for the reaction: 2NO(g) + Cl2(g) → 2NOCl(g) They measure the concentrations of all the reactants and products at different times and obtain the following data:Time (s) | [NO] (M) | [Cl2] (M) | [NOCl] (M):--: | :--: | :--: | :--:0 | 0.100 | 0.200 | 0.00010 | 0.080 | 0.180 | 0.04020 | 0.064 | 0.160 | 0.06430 | 0.051 | 0.140 | 0.08140 | 0.041 | 0.120 | 0.089What is the rate constant (k) for this reaction?

Question

A chemistry student wants to determine the rate constant for the reaction: 2NO(g) + Cl2(g) → 2NOCl(g) They measure the concentrations of all the reactants and products at different times and obtain the following data:Time (s) | [NO] (M) | [Cl2] (M) | [NOCl] (M):--: | :--: | :--: | :--:0 | 0.100 | 0.200 | 0.00010 | 0.080 | 0.180 | 0.04020 | 0.064 | 0.160 | 0.0

Answer 1

To determine the rate constant  k  for this reaction, we first need to find the rate law for the reaction. The general form of the rate law is:Rate = k[NO]^x[Cl2]^yWe need to find the values of x and y  the orders of the reaction with respect to NO and Cl2, respectively  by analyzing the data provided.Let's compare the data between time 0s and 10s:Rate1 = k[0.100]^x[0.200]^yRate2 = k[0.080]^x[0.180]^yDivide Rate1 by Rate2: Rate1/Rate2  =  [0.100]^x[0.200]^y  /  [0.080]^x[0.180]^y Now, we need to find the rate of the reaction. We can do this by looking at the change in concentration of NOCl over time. Between 0s and 10s, the concentration of NOCl increased by 0.040 M. Since the stoichiometry of the reaction is 2NO + Cl2  2NOCl, the rate of the reaction with respect to NOCl is half the rate with respect to NO:Rate1/Rate2 =  0.040/10  /  0.020/10  = 2Now we have:2 =  [0.100]^x[0.200]^y  /  [0.080]^x[0.180]^y To simplify this equation, we can take the logarithm of both sides:log 2  = x log 0.100  - log 0.080   + y log 0.200  - log 0.180  0.301 = 0.124x + 0.041yNow, let's compare the data between time 10s and 20s:Rate3 = k[0.064]^x[0.160]^yDivide Rate2 by Rate3: Rate2/Rate3  =  [0.080]^x[0.180]^y  /  [0.064]^x[0.160]^y Between 10s and 20s, the concentration of NOCl increased by 0.024 M. The rate of the reaction with respect to NOCl is:Rate2/Rate3 =  0.024/10  /  0.020/10  = 1.2Now we have:1.2 =  [0.080]^x[0.180]^y  /  [0.064]^x[0.160]^y Take the logarithm of both sides:log 1.2  = x log 0.080  - log 0.064   + y log 0.180  - log 0.160  0.079 = 0.096x + 0.054yNow we have a system of two linear equations:0.301 = 0.124x + 0.041y0.079 = 0.096x + 0.054ySolving this system of equations, we get:x  1y  1So, the rate law is:Rate = k[NO][Cl2]Now we can find the rate constant  k  using the data from any of the time points. Let's use the data from time 10s:Rate2 = k[0.080][0.180]The rate of the reaction with respect to NOCl at 10s is 0.020 M/s. Since the stoichiometry of the reaction is 2NO + Cl2  2NOCl, the rate of the reaction with respect to NO is half the rate with respect to NOCl:Rate2 = 0.020 M/sNow we can solve for k:0.020 = k 0.080  0.180 k  1.39 M^-1 s^-1 The rate constant  k  for this reaction is approximately 1.39 M^-1 s^-1 .