Chemistry Forum

The rate constant of the elementary reaction A₂+B₂=2AB is equal to 0.161 mol^-1*s^-1. The initial concentrations of the reactants are A₂=0.04 mol/l and B₂ 0.05 mol/l. Calculate the initial rate of the reaction. (I made this request and it turned out that the speed of the reaction = 3.2*10^-4 mol/l*s.)
b) Calculate the speed in the cast when A₂ is reduced 2 times.(Here the solution to the exercise is different and the way I solved it is different. Since the concentration of A₂ has been reduced by 2 times, then it falls to 0.02. So it turns out that during this reaction 0.02 mol of A₂ was consumed. Since the ratio is 1:1, 0.02 mol of B₂ should be consumed. Therefore, the concentration of B₂ falls to 0.03. Everything says that since A₂ is reduced by 2 times and B₂ is reduced by 2 times and the concentration of B₂ is 0.025. I don't understand why?)

I think its not a consumption.
The concentration is reduced from 0,04 to 0,02
The reaction took place 0,04 with 0,05 means to keep the ratio it is 0,02 to 0,025. B is also reduced 2 times from 0,05 to 0,025.

yes but the consumption of A₂ is 0.02

It said A₂ is reduced 2 times, not consummed. So it was taken out or removed. The same has to be done with B₂.
The reaction is of course 1:1.
But if 0.02 mol A₂ has reacted, then also 0,02 mol B₂ has reacted and the new value is 0,03 mol.

Okay thxx a lott

Oh, I got some useful infor here. Big thanks

To calculate the initial rate of the reaction, we can use the rate law of a first-order reaction:

Velocity = k[A₂][B₂]

where k is the rate constant and [A₂] and [B₂] are the initial concentrations of the reactants.

a) To calculate the initial rate of the reaction:
k = 0.161 mol^-1*s^-1
[A₂] = 0.04 mol/l
[B₂] = 0.05 mol/l

Initial velocity = k[A₂][B₂]
= 0.161 mol^-1s^-1 * 0.04 mol/l * 0.05 mol/l
= 3.2 * 10^-4 mol/ls

So, the initial rate of the reaction is 3.2 * 10^-4 mol/l*s.

b) To calculate the rate of the reaction when the concentration of A₂ is halved, we must use the new concentrations:

[A₂] = 0.02 mol/l
[B₂] = ?

In the stoichiometric reaction, the ratio is 1:1, which means that for every mole of A₂ that is consumed, one mole of B₂ is also consumed.

If the concentration of A₂ is halved (0.02 mol/l), then 0.02 mol of B₂ will also be consumed.

To find the new concentration of B₂, we must subtract the 0.02 mol consumed from the initial concentration of B₂ (0.05 mol/l):

[B₂] = 0.05 mol/l - 0.02 mol
= 0.03 mol/l

Therefore, the new concentration of B₂ is 0.03 mol/L.

The difference in the answer is due to an error in your calculation. The B₂ concentration is reduced to 0.03 mol/l, not 0.025 mol/l. Following the stoichiometry of the reaction, 0.02 mol of B₂ is consumed when the concentration of A₂ is halved, resulting in a final concentration of 0.03 mol/L for B₂.

It's really thanks to everyone that I can do more chemistry homework. Balancing moles of things. I'm really confused and don't understand how to do it

Give an example, what is your problem?