Question

The reaction A + 2B + C $\to$ 2D + E is found to be 1, 2 and Zero order with respect to A,B and C respectively. What will be the final rate, if concentration of each reactant is doubled?

• A. 2 times
• B. 4 times
• C. 8 times
• D. 16 times

Answer 1

Answer: (C)

8 times

The given reaction is:

$A+2B+C⟶2D+E$

According to question,

Rate=$K\left[A{]}^1\right[B{]}^2[C{]}^0$ $-\left(i\right)$

where, $K$ is rate constant

Now, if the concentration of each reactant is doubled then,

$(Rate{)}_{new}$=$K\left[2A{]}^1\right[2B{]}^{[}2C{]}^0$

=$K.2\left[A{]}^1.4\right[B{]}^2.2[C{]}^0$

$(Rate{)}_{new}=8.K[A{]}^1\left[B{]}^2\right[C{]}^0$ $-\left(ii\right)$

Now, Dividing $\left(ii\right)$ by $\left(i\right)$ :-

$\frac{(Rate{)}_{new}}{Rate}=8$

So, the final rate will be $8$ times to that of initial.