if the concentration of all the reactants doubled, what works happen to the rate
I know the reaction rate would increase, but in a previous example it asked the question and the answer was that the rate increases by 16 and I can't figure out how much it would increase for this question
The rate law expression for this question is R = K [ A ] ^ 1 [ B ] ^ 1
You plug the numbers into the rate expression.
rate = k*A*B
If you double A and double B then rate goes up by 4. Or the longer way is to make up a number for k, A and B. Let's make them easy and call k = 1, A = 2 and B = 3 so rate1 = 1*2*3 = 6
Now lets double A and double B so
rate = 1*4*6 = 24. Did the rate change by a factor of 4? yes, from 6 to 24 which is 4x more.
To determine how the rate would change when the concentration of all reactants is doubled, we can use the rate law expression provided: R = K [ A ]^1 [ B ]^1.
According to the rate law, the rate of the reaction is proportional to the concentrations of the reactants raised to their respective powers. In this case, both reactants A and B have an exponent of 1 in the rate law expression.
If the concentration of each reactant is doubled, it means their initial concentrations have now become 2[A] and 2[B]. Let's calculate the rate before and after doubling the concentration:
Initial rate: R₁ = K [ A ]^1 [ B ]^1
Final rate: R₂ = K [ 2[A] ]^1 [ 2[B] ]^1
= K (2[A])(2[B])
= 4K[A][B]
Comparing the final rate (R₂) to the initial rate (R₁), we see that the rate has increased by a factor of 4. So, in this case, if the concentration of all the reactants is doubled, the rate of the reaction would increase by a factor of 4 or would be four times faster than the initial rate.