The rate of reactions
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The rate of reactions
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?)
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?)
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Re: The rate of reactions
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.
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.
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Re: The rate of reactions
yes but the consumption of A2 is 0.02
- ChenBeier
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Re: The rate of reactions
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.
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.
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Re: The rate of reactions
Okay thxx a lott
Re: The rate of reactions
Oh, I got some useful infor here. Big thanks
- FRancoFrias
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Re: The rate of reactions
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.
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.
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Re: The rate of reactions
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
- ChenBeier
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Re: The rate of reactions
Give an example, what is your problem?