Question

The reaction $A + 2B + C \to 2D + E$ is found to be 1, 2 and zero order with respect to A, B, 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

Hint:
The rate of reaction is the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or concentration of a reactant that is consumed in a unit of time.
Formula used:
Rate\[\; = {\text{ }}k{\text{ }}{\left[ A \right]^a}{\left[ B \right]^b}\]
Where, k= rate constant
A and B are concentrations in mol \[\]
a is order of reaction with respect to A
b is order of reaction with respect to B

Complete step by step answer:
The given reaction $A + 2B + C \to 2D + E$
Now according to the rate formula,
Rate\[ = {\text{ }}K{\text{ }}{\left[ A \right]^1}{\left[ B \right]^2}{\left[ C \right]^0}\]...............................................(i)
Now, if the concentration of each reactant is doubled, then we will get
New Rate\[ = {\text{ }}k\left[ {2A} \right]^1{\text{ }}\left[ {2B} \right]^2{\text{ }}\left[ {2C} \right]^0\]
\[ = {\text{ }}K.2{\text{ }}\left[ A \right]^1.4\left[ B \right]^2\left[ C \right]^0\]
\[ = 8.K\left[ A \right]^1\left[ B \right]^2\left[ C \right]^0\]………………………………………………….(ii)
Now dividing (ii) by (i)
We get, \[\dfrac{{New\;Rate}}{{Rate}} = 8\]
Therefore, the final rate will be 8 times that of initial.

Hence, option C is correct.

Note:
The factors that affect the rate of reaction are surface area of a solid reactant, temperature, nature of reactant, presence/absence of catalyst, concentration or pressure of a reactant.