The following general reaction was found to be first-order with respect to A and second-order with respect to B:
A + B ---> products
What would the rate of this reaction be in terms of k in a one liter vessel containing 4 moles of A and 7 moles of B?
rate = k(A)(B)^2
rate = k(4)(7)^2
rate = k*4*49
you finish
To determine the rate of the reaction, we need to use the rate law equation. For a first-order reaction with respect to A and a second-order reaction with respect to B, the rate law equation would be:
Rate = k[A][B]^2
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of A
- [B] is the concentration of B
In this case, we know the reaction is first-order with respect to A and second-order with respect to B. Therefore, the rate law equation becomes:
Rate = k[A][B]^2
Given that we have 4 moles of A and 7 moles of B in a 1 liter vessel, we need to convert the moles to concentrations.
The concentration of A ([A]) can be calculated as:
[A] = moles of A / volume of the vessel
[A] = 4 moles / 1 liter = 4 M
The concentration of B ([B]) can be calculated as:
[B] = moles of B / volume of the vessel
[B] = 7 moles / 1 liter = 7 M
Now, we can substitute these values into the rate law equation:
Rate = k[A][B]^2
Rate = k(4 M)(7 M)^2
Rate = k(4)(49) M^3
Therefore, the rate of the reaction in terms of k would be k(196) M^3.