Question

A reaction in which A, B, and C react to form products is zero order in A, one-half order in B, and second order in C.

Answer 1

Answer:

a. $r=k[B]^{1/2}[C]^2$

b. $Order=3.5$

c. It does not affect the rate.

d. $\sqrt{2}$.

e. 4.

f. 4$\sqrt{2}$.

Explanation:

Hello,

In this case, considering the given information, we have:

a. $r=k[B]^{1/2}[C]^2$

b. By adding 1/2 and 2 (powers for B and C), the overall order is:

$Order=\frac{1}{2} +2\\\\Order=3.5$

c. It is not changed, since the concentration of A is not affecting the rate due to its specific zeroth-order.

d. In this case, if the concentration of B is doubled, such term in the equation shows:

$\sqrt[n]{x} \frac{r_f}{r_i}=\frac{[2B]^{(1/2)} [C]^2}{[B]^{(1/2)}[C]^2} \\\\\frac{r_f}{r_i}=\frac{[2B]^{(1/2)}}{[B]^{(1/2)}}\\\\\frac{r_f}{r_i}=(\frac{2}{1})^{1/2}\\\\r_f=\sqrt{2} r_i$

It means that the rate increases by a factor of $\sqrt{2}$.

e. In this case, if the concentration of C is doubled, such term in the equation shows:

$\frac{r_f}{r_i}=\frac{[B]^{(1/2)}[2C]^2}{[B]^{(1/2)}[C]^2} \\\\\frac{r_f}{r_i}=(\frac{[2C]}{[C]})^{2}\\\\\frac{r_f}{r_i}=(\frac{2}{1})^{2}\\\\r_f=4r_i$

It means that the rate increases by a factor of 4.

f. In this case, if the concentration of both B and C are doubled, such terms in the equation shows:

$\frac{r_f}{r_i}=\frac{[2B]^{(1/2)}[2C]^2}{[B]^{(1/2)}[C]^2} \\\\\frac{r_f}{r_i}=\frac{2^{(1/2)}2^2}{1^{(1/2)}1^2} \\\\\frac{r_f}{r_i}=4\sqrt{2} \\\\r_f=4\sqrt{2} r_i$

It means that the rate increases by a factor of 4$\sqrt{2}$.

Best regards.