Physical Chemistry- Rates and Kinetics

AI Thread Summary
The discussion revolves around calculating the rate of a chemical reaction involving nitric oxide and chlorine, with the reaction being second order in NO and first order in Cl2. The initial rate is given as 2.4x10-3 moldm-3sec-1, leading to the determination of the rate constant k as 0.000384. When half of the chlorine reacts, the concentration of Cl2 is 0.5M, but the stoichiometry indicates that for every mole of Cl2 consumed, two moles of NO are used, affecting the concentration of NO. Participants clarify that the concentration of NO must be recalculated based on the stoichiometric relationship rather than simply halving it, leading to a final rate of 0.000192 moldm-3sec-1 when considering the correct concentrations. The importance of accurate stoichiometric calculations in determining reaction rates is emphasized throughout the discussion.
chanderjeet
Messages
24
Reaction score
0

Homework Statement



The Reaction; 2NO(g) + Cl2 (g) → 2NOCl (g)

Is second order in NO and first order in Cl2. Five moles of nitric oxide and two moles of Cl2 were brought together in a volume of 2dm3 and initial rate was 2.4x10-3 moldm-3sec-1. What will be the rate when one half of the chlorine has reacted?

Homework Equations



Rate = k [NO]2 [Cl2]


The Attempt at a Solution



Initial Rate= 2.4x10-3 moldm-3sec-1

2.4x10-3 = k [NO]2 [Cl2]

5mols of NO in 2dm3
= 5mols in 2L
∴ 1L contains 2.5mols
[NO] = 2.5M

2mols of Cl2 in 2dm3
= 2mols in 2L
∴ 1L contains 1mol
[Cl2] = 1M

2.4x10-3 = k (2.5)2 (1)
k = 2.4x10-3/6.25
= 0.000384

When half of the chorine is used (1mol), molarity = 0.5M

∴ Rate = (0.000384) (2.5)2 (0.5)
Rate = 0.0012moldm-3sec-1

Am I going about this the right way or this completely wrong?
 
Physics news on Phys.org
Close, but no cigar yet.

How much NO was consumed together with the chlorine?
 
Wait, sorry I read the question wrong. Don't mind me, listen to Borek.
 
Borek said:
Close, but no cigar yet.

How much NO was consumed together with the chlorine?

So does that mean that since half the Cl2 (0.5mols) is used then half the NO is used (2.5mols)?

[NO] 2.5mols in 2dm3
2.5mols in 2L
∴ 1L = 1.25mols
[NO] = 1.25M

Rate = (0.000384) (1.25)2 (0.5)
= 0.0003moldm-3sec-1
 
chanderjeet said:
So does that mean that since half the Cl2 (0.5mols) is used then half the NO is used (2.5mols)?

No, it is not about "half". It is a trivial stoichiometry.
 
Borek said:
No, it is not about "half". It is a trivial stoichiometry.

From the balanced equation, a 2: 1 mol ratio is noted between NO and Cl2

So if one mol of Cl2 is used up then 2 mols NO would be used up

[NO] = 1M
[Cl2] = 0.5M

Rate= (0.000384) (1)2 (0.5)
= 0.000192moldm-3s-1
 
No.

Please pay attention to what you are doing, you have made a second careless mistake. You were right up to 2 moles of NO being consumed.
 
Borek said:
No.

Please pay attention to what you are doing, you have made a second careless mistake. You were right up to 2 moles of NO being consumed.

I thought the volume was given so that the concentration can calculated an utilized in the rate expression. Am I not supposed to calculate the concentration based on the mols being consumed? :confused:
 
Using your logic once all chlorine is consumed, and there is no more chlorine present, its concentration is 1 M, as 2 moles were consumed. So all was consumed, but the concentration has the maximum value. Something doesn't sound right.
 
Back
Top