the reaction A+3B—>2C+D is first order with respect to reactant A and second order with respect to reactant B if the concentration of A is doubled and the concentration of B is halved the rate of the reaction would---------------by a factor of-----------
rate=constant*a^2 ( b^3)
= constant*4* 1/8= 1/2 rate
oops, thanks dr bob
rate=constant*a*b^3=1/4
if B is halved it should be 1.5 not 1/8 right
To determine how the rate of the reaction changes when the concentrations of reactants A and B are altered, we need to consider the rate equation for the reaction.
The rate equation for a reaction is typically of the form:
rate = k[A]^x[B]^y
Where:
- rate is the rate of the reaction
- k is the rate constant
- [A] and [B] are the concentrations of reactants A and B, respectively
- x and y are the orders of the reaction with respect to reactants A and B, respectively
Given that the reaction A+3B → 2C+D is first order with respect to reactant A and second order with respect to reactant B, we have:
rate = k[A]^1[B]^2
Now, let's consider what happens when the concentration of A is doubled and the concentration of B is halved.
If the concentration of A is doubled, we can rewrite the rate equation as:
rate2 = k(2[A])^1[B]^2 = 2k[A][B]^2
If the concentration of B is halved, we can rewrite the rate equation as:
rate3 = k(2[A])(0.5[B])^2 = 0.5k[A][B]^2
Comparing rate2 and rate3 to the original rate equation (rate1 = k[A][B]^2), we can determine how the rate changes:
rate2 / rate1 = (2k[A][B]^2) / (k[A][B]^2) = 2
rate3 / rate1 = (0.5k[A][B]^2) / (k[A][B]^2) = 0.5
Therefore, when the concentration of A is doubled and the concentration of B is halved, the rate of the reaction would increase by a factor of 2 and then decrease by a factor of 0.5.