Question

question 23 (1 point)
a reaction has the following rate law:
r = ka^2b^0
what would be the effect on the rate if the concentration of both a and b were
doubled?
the rate would double
the rate would quadruple
the rate would decrease by half
the rate would increase in an unpredictable way
the rate would increase by a factor of eight
question 24 (1 point)

Explanation:

Step1: Analyze the original rate law

The original rate law is \(r = k[A]^2[B]^0\). Since \([B]^0 = 1\), the original rate is \(r_1=k[A]^2\).

Step2: Substitute the new concentrations

When \([A]\) becomes \(2[A]\) and \([B]\) becomes \(2[B]\), the new rate \(r_2=k(2[A])^2(2[B])^0\). Because \((2[B])^0 = 1\), then \(r_2 = k\times4[A]^2\times1\).

Step3: Compare the new and original rates

We know \(r_1 = k[A]^2\) and \(r_2=4k[A]^2\). So, \(\frac{r_2}{r_1}=\frac{4k[A]^2}{k[A]^2}=4\).

Answer:

The rate would quadruple.