Question

For the chemical reaction A+B+C→E, the rate of the reaction is doubled when the concentration of B was doubled, and when the concentration of both A and B was doubled rate became doubled and when the concentration of both B and C was doubled rate became quadrupled. What is the order with respect to A, B and C and the total order?
A. 0, 1, 2; 3
B. 1, 1, 0; 2
C. 0, 1, 1; 2
D. 1, 0, 2; 3

Answer 1

Hint: The reaction rate is the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

Complete answer:
- The reaction rate is always defined as the change in the concentration divided by the change in time, with an extra term that is 1 divided by the stoichiometric coefficient.
- For the given reaction, A+B+C→E
The rate of the reaction will be,${{r}_{1}}=k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}$(1)
Where,
x = order w.r.t A
y = order w.r.t B
z = order w.r.t C
Now, let us mention the given facts
$\begin{align}
  & {{r}_{2}}=2{{r}_{1}}=k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
 & {{r}_{2}}={{2}^{y}}.k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
\end{align}$ (2)
$\begin{align}
  & {{r}_{3}}=2{{r}_{1}}=k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
 & {{r}_{3}}={{2}^{x}}{{.2}^{y}}k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
\end{align}$(3)
$\begin{align}
  & {{r}_{4}}=4{{r}_{1}}=k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
 & {{r}_{4}}={{2}^{y}}{{.2}^{z}}k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
\end{align}$ (4)
Now, on dividing eq. 2 by 3, we get
\[\begin{align}
  & \dfrac{{{r}_{2}}}{{{r}_{3}}}=\dfrac{2{{r}_{1}}}{2{{r}_{1}}}=\dfrac{{{2}^{y}}.k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}}{{{2}^{x}}{{.2}^{y}}k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}} \\
 & 1={{2}^{x}} \\
 & {{2}^{0}}={{2}^{x}} \\
 & x=0 \\
\end{align}\]
$\begin{align}
  & \dfrac{{{r}_{2}}}{{{r}_{4}}}=\dfrac{2{{r}_{1}}}{4{{r}_{1}}}=\dfrac{{{2}^{y}}.k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}}{{{2}^{y}}{{.2}^{z}}.k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}}} \\
 & \dfrac{1}{2}=\dfrac{1}{{{2}^{z}}} \\
 & {{2}^{z}}={{2}^{1}} \\
 & z=1 \\
\end{align}$
$\begin{align}
  & {{r}_{2}}=2{{r}_{1}} \\
 & k{{[A]}^{x}}{{[2B]}^{y}}{{[C]}^{z}}=2k{{[A]}^{x}}{{[B]}^{y}}{{[C]}^{z}} \\
 & {{2}^{y}}{{[B]}^{y}}=2{{[B]}^{y}} \\
 & {{2}^{y}}={{2}^{1}} \\
 & y=1 \\
\end{align}$
Thus,
x = 0
y = 1
z = 1
So, $rate=k{{[A]}^{0}}{{[B]}^{1}}{{[C]}^{1}}$
And, the order of the reaction = 0+1+1=2

Therefore, the answer to the question is (C) 0, 1, 1; 2

Note:
Reaction rate determines how fast or how slow the reaction is. Reaction time is the time taken to complete a reaction to a certain extent. If the reaction rate is high for a particular reaction, then the reaction time is low. Also, if the reaction rate is low, then the reaction time will be longer.