The reaction 2NO(g) + Cl2(g) --> 2NOCl(g) obeys the rate law rate = k [NO]2 [Cl2]....

The reaction 2NO(g) + Cl₂(g) --> 2NOCl(g) obeys the rate law rate = k [NO]² [Cl₂]. The following mechanism is proposed:

NO (g) + Cl₂ (g) --> NOCl₂(g)

NOCl₂(g) + NO (g) -->2NOCl (g)

a) What would the rate law be if the first step was rate determining?

b) Based on the observed rate law, what can be concluded about the relative rates of the 2 reactions?

Expert Solution

a.

Rate laws must be determined experimentally, however if the first step is a bimolecular elementary step, then the rate law would be: rate = k[NO]^2 [Cl2].

b.

The observed rate law appears to be consistent with the first step being fast and the second step being slow. This can be proven as follows:

Equation 1) NO + Cl2 <---> NOCl2 (fast)

Equation 2) NOCl2 + NO --> 2NOCl (slow)
2NO + Cl2 --> 2NOCl (overall equation)

rate = k2 [NOCl2] [NO] The slow step determines the rate law, however intermediates cannot be in the rate law, so we have to substitute for it.

The fast step is at equilibrium, so the rate in the forward reaction is equal to the rate in the reverse rxn.
k1 [NO] [Cl2] = k-1 [NOCl2]

Solve for the intermediate concentration.

[NOCl2] = k1/k-1 [NO] [Cl2] (substitute into the rate law)
rate = k2k1/k-1 [NO]^2 [Cl2]

The assumption that the fast and slow steps are what they are is because it matches the experimental rate law.

rate = k2 [NOCl2] [NO]
rate = k[NO]^2 [Cl2]

queen_honey_blossom answered 9 months ago

For the reaction 2NO(g) + Cl2(g) --->2NOCl(g) if the concentration of NO is tripled, the rate...

For the reaction 2NO(g) + Cl2(g) --->2NOCl(g) if the concentration of NO is tripled, the rate of the reaction increases by a factor of nine. if the concentration of Cl2 is cut in half, the rate of the reaction is decreased to half the original rate. Find the order of reaction for each reactant and write the rate expression for the reaction.

At 35 degrees C, Kc = 1.6x10^-5 for the reaction: 2NOCl(g),<--> 2NO(g) + Cl2(g) Calculate the...

At 35 degrees C, Kc = 1.6x10^-5 for the reaction: 2NOCl(g),<--> 2NO(g) + Cl2(g) Calculate the concentrations of all species at equilibrium for each of the following origional mixtures. a. 1.00 mol NOCl in a 1.0 L container b. 1.00 mol NO and 0.50 mol Cl2 in a 1.0 L container c. 1.00 mol NO and 0.75 mol Cl2 in a 1.0 L container The answers are: a. [NOCl]= 1.0M, [NO]= 0.032M, [Cl2]= 0.016M; b. [NOCl]= 1.0M, [NO]= 0.032M, [Cl2]=...

At 35oC, K = 1.6 x 10-5 for the following reaction 2NOCl(g) <----> 2NO(g) + ...

At 35oC, K = 1.6 x 10-5 for the following reaction 2NOCl(g) <----> 2NO(g) + Cl2(g) Calculate the concentration of all species at equilibrium for each of the following original mixtures. 1) 3.0 moles of pure NOCl in a 2.0 L flask 2) 2.0 moles of NOCl and 2 moles of NO in a 3 L flask 3) 3.0 moles of NOCl 1 mole of Cl2 in a 2.0 L flask Show work please

If delta H rxn is -75.2 kJ, 2NO(g) + Cl2 (g) <---> 2NOCl(g)

If delta H rxn is -75.2 kJ, 2NO(g) + Cl2 (g) <---> 2NOCl(g) A scientist places four moles of nitrogen monoxide and two moles of chlorine gas into flask A and the same amounts into Flask B, then allows the systems to reach equilibrium. Flask A is at 25 degrees Celsius and Flask B is at 200 degrees Celsius A) In which flask will the reaction occur faster? Explain. B) In which flask will the reaction occur to a greater...

Consider the following equilibrium: 2NOCl(g) 2NO(g) + Cl2(g)with K = 1.6 × 10–5. 1.00 mole of pure...

Consider the following equilibrium: 2NOCl(g) 2NO(g) + Cl2(g)with K = 1.6 × 10–5. 1.00 mole of pure NOCl and 1.50 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g).

The reaction 2NOCl(g)=2NO(g)+Cl2(g) comes to equilibrium at 1 bar total pressure and 227C when the partial...

The reaction 2NOCl(g)=2NO(g)+Cl2(g) comes to equilibrium at 1 bar total pressure and 227C when the partial pressure of the nitrosyl chloride, NOCl, is 0.64 bar. Only NOCl was present initially. (a) Calculate deltarGo for this reaction. (b) At what total pressure will the partial pressure of Cl2 be at 0.1 bar?

Q1) part A: Consider the decomposition of NOCl: 2NOCl(g) → 2NO(g) + Cl2(g) During the decomposition,...

Q1) part A: Consider the decomposition of NOCl: 2NOCl(g) → 2NO(g) + Cl2(g) During the decomposition, the concentration of cyclopropane is measured. By graphing [NOCl] vs. time, ln[NOCl] vs. time and 1/[NOCl] vs. time, the plot of 1/[NOCl] vs. time gives a straight line with a slope of 9.98 x 10-2 at 129oC. (a) What is the reaction order for the decomposition of NOCl. (b) Calculate the rate constant of the reaction. (c) If the initial concentration of NOCl(g) is...

Question 11 5 pts Nitrosyl chloride, NOCl, decomposes to NO and Cl2. 2NOCl (g) ----> 2NO...

Question 11 5 pts Nitrosyl chloride, NOCl, decomposes to NO and Cl2. 2NOCl (g) ----> 2NO (g) + Cl2 (g) Determine the overall order and rate constant "k" for this reaction from the following data: [NOCl] (M) 0.10 0.20 0.30 Rate (M/h) 8.0 x 10–10 3.2 x 10–9 7.2 x 10–9

The reaction 2A + B → C obeys the rate law –rB = kCB2, where k...

The reaction 2A + B → C obeys the rate law –rB = kCB2, where k = 0.25 L/mol.s at 75˚C. A is available as a 1.5 M solution and B as a 1 M solution. a) A 26 L/s stream of solution A and 30 L/s stream of solution B are combined just before being introduced into a CSTR. If a 150 L reactor is available, how many mole/h of C could be produced? b) You receive some financing...

For the following reaction: 2HI (g) ----> H2(g) + I2(g), the rate law is rate= k[HI]^2....

For the following reaction: 2HI (g) ----> H2(g) + I2(g), the rate law is rate= k[HI]^2. You are starting with 1.0M HI. How long will the first half-life reaction be? (The professor adds: Note that since you don't know the value for the rate constant k, just leave it in your answers.)

Question