Rate of a Reaction - Definition, Representation, Practice Problems and FAQs

Can we convert graphite into a diamond? Well, yes, according to thermodynamics, graphite can be converted to diamond as this reaction is feasible but it will take millions of years to convert graphite into diamond. So, actually, it's not a great idea to sit back and wait for a few million years. We actually need to understand ways to pace up or slow down the reaction, so that we can get the desired result according to our needs.

Please enter alt text

To make it possible, first, we need to understand what is the “speed of reaction” or in other words the ‘rate of a reaction.

Table of contents

What is the Rate of a Reaction?
Representation of Rate of a Reaction in a Chemical Reaction
Practice Problems
Frequently Asked Questions - FAQs

What is the Rate of a Reaction?

The rate of a reaction is defined as the change in concentration of the reactants or products per unit time. Basically, it is the speed at which reactants are converted into products.

Representation of Rate of a Reaction in a Chemical Reaction

Consider a reaction, $R \to P,$

The rate of reaction for the above chemical reaction can be expressed in two ways:

(i) The rate of decrease in the concentration of reactant
(ii) The rate of increase in the concentration of product

Rate of disappearance of R = $\frac{D e c r e a s e i n c o n c e n t r a t i o n o f R}{T i m e t a k e n}$
Rate of appearance of P = $\frac{I n c r e a s e i n c o n c e n t r a t i o n o f P}{T i m e t a k e n}$

Rate of reaction (in terms of reactant, R) = $- \frac{1}{S t o i c h i o m e t r i c c o e f f i c i e n t} (\frac{C h a n g e i n c o n c e n t r a t i o n o f R}{T i m e t a k e n})$
Rate of reaction (in terms of product, P) = $\frac{1}{S t o i c h i o m e t r i c c o e f f i c i e n t} (\frac{C h a n g e i n c o n c e n t r a t i o n o f P}{T i m e t a k e n})$

Note: The negative sign is used to indicate that the concentration of reactant is decreasing.

Consider a chemical reaction, aA(g) + bB(g) cC(g) + dD(g)

Rate of a reaction can be expressed in the below mentioned way:

Rate of a reaction = $- \frac{1}{a} \frac{d [A]}{d t} = - \frac{1}{b} \frac{d [B]}{d t} = \frac{1}{c} \frac{d [C]}{d t} = \frac{1}{d} \frac{d [D]}{d t}$

Where,

Rate of disappearance of A = $- \frac{d [A]}{d t} o r - \frac{{d P}_{A}}{d t}$
Rate of disappearance of B = $- \frac{d [B]}{d t} o r - \frac{{d P}_{B}}{d t}$
Rate of appearance of C = $\frac{d [C]}{d t} o r \frac{{d P}_{C}}{d t}$
Rate of appearance of D = $\frac{d [D]}{d t} o r \frac{{d P}_{D}}{d t}$

Unit of the rate of reaction is mol L^-1 s^-1 and for reaction in the gaseous phase is atm s^-1.

Recommended video

Practice Problems

Q1. For a reaction, $\frac{1}{2} A \to 2 B,$ the relation between the rate of disappearance of ‘A’ and the rate of appearance of ‘B’ can be expressed by:

$(A) - \frac{d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t} (B) - \frac{d [A]}{d t} = \frac{d [B]}{d t} (C) - \frac{d [A]}{d t} = 4 \frac{d [B]}{d t} (D) - \frac{d [A]}{d t} = \frac{1}{2} \frac{d [B]}{d t}$

Answer: (A)

Solution:

Rate of reaction = $- \frac{1}{\frac{1}{2}} \frac{d [A]}{d t} = \frac{1}{2} \frac{d [B]}{d t}$

⇒ Rate of reaction = $- 2 \frac{d [A]}{d t} = \frac{1}{2} \frac{d [B]}{d t}$ …………(i)

Rate of disappearance of ‘A’ = $- \frac{d [A]}{d t}$

Rate of appearance of ‘B’ = $\frac{d [B]}{d t}$

Dividing equation (i) by 2, we get;

$- \frac{d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}$

Q2. For the reaction, $2 A + 3 B + \frac{3}{2} C \to 3 P$ , which statement is correct?

$(A) \frac{d [A]}{d t} = \frac{3}{2} \frac{d [B]}{d t} = \frac{3}{4} \frac{d [C]}{d t} (B) \frac{d [A]}{d t} = \frac{d [B]}{d t} = \frac{d [C]}{d t} (C) \frac{d [A]}{d t} = \frac{2}{3} \frac{d [B]}{d t} = \frac{4}{3} \frac{d [C]}{d t} (D) \frac{d [A]}{d t} = \frac{2}{3} \frac{d [B]}{d t} = \frac{3}{4} \frac{d [C]}{d t}$

Answer: (C)
Solution: Rate of a reaction $= - \frac{1}{2} \frac{d [A]}{d t} = - \frac{1}{3} \frac{d [B]}{d t} = - \frac{1}{\frac{3}{2}} \frac{d [C]}{d t}$
Multiplying above equation by -2, we get;
$\frac{d [A]}{d t} = \frac{2}{3} \frac{d [B]}{d t} = \frac{4}{3} \frac{d [C]}{d t}$

Q3 of rate is:

(A) mol L^-1s^-1
(B) mol L^-1s
(C) M s^-1
(D) Both (A) and (C)

Answer: (D)

Solution:

Rate of reaction = $\frac{1}{S t o i c h i o m e t r i c c o e f f i c i e n t} (\frac{C h a n g e i n c o n c e n t r a t i o n}{T i m e t a k e n})$

$= \frac{m o l L^{- 1} o r M}{s} = m o l L^{- 1} s^{- 1} o r {M s}^{- 1}$

Q4. For the reaction: 4NH₃ + 5O₂--> 4NO + 6H₂O

$\frac{d [N O]}{d t} = 4 .8 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

The rate of formation of water is:

(A) 7.2 x 10^-2 mol L^-1s^-1
(B) 1.2 x 10^-4 mol L^-1s^-1
(C) 2.88 x 10^-3 mol L^-1s^-1
(D) 7.2 x 10^-4 mol L^-1s^-1

Answer: (D)

Solution:

Rate of reaction = $- \frac{1}{4} \frac{d [N H_{3}]}{d t} = - \frac{1}{5} \frac{d {[O}_{2}]}{d t} = \frac{1}{4} \frac{d [N O]}{d t} = \frac{1}{6} \frac{d {[H}_{2} O]}{d t}$

$\Rightarrow \frac{1}{4} \frac{d [N O]}{d t} = \frac{1}{6} \frac{d {[H}_{2} O]}{d t}$

$\frac{d {[H}_{2} O]}{d t} = \frac{1}{4} \times 6 \times 4 .8 \times 10^{- 4} = 7 .2 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

Frequently Asked Questions-FAQs

Q1. What does the negative sign indicate in the rate of reaction?

Answer: Consider a reaction, $R \to P$

The rate of reaction in terms of reactant = $- \frac{d [R]}{d t}$

As the concentration of the reactant is decreasing, that's why the negative sign is used.

Q.2 What is the unit of rate of reaction in the gaseous phase?

Answer:

Consider a reaction, $A (g) \to B (g)$

Rate of a reaction = $\frac{C h a n g e i n p r e s s u r e}{T i m e t a k e n} = \frac{a t m}{s}$
The unit of rate of a reaction in the gaseous phase is atm s^-1.

Q3. How do we measure the speed of a reaction in the laboratory?

Answer:

The speed of the reaction can be measured by calculating the rate of a reaction.

In the laboratory, it can be measured by noting the concentration of the reaction medium at different intervals of time. Later rate of reaction can be calculated using the formula:

$R a t e = \frac{T o t a l c h a n g e i n c o n c e n t r a t i o n}{T o t a l t i m e t a k e n} = |\frac{C_{f i n a l} - C_{i n i t i a l}}{t_{f i n a l} - t_{i n i t i a l}}| = \frac{△ c}{△ t}$

Q4. What is the role of the stoichiometric coefficient in the rate of a chemical reaction?

Answer:

Consider a chemical reaction, $A + 2 B \to P$

Let's say, if 1 mol of A and 2 mol of B are consumed simultaneously in five seconds, the rate of consumption for A and B will be different. When 1 mol of A is consumed, in those five seconds, in same five seconds actually 2 mol of B are being consumed.

Rate of disappearance of A ≠ Rate of disappearance of B

In order to relate the rate of disappearance of reactants, we define the rate of a reaction. Such that we divide the rate of disappearance of each reactant by their corresponding stoichiometric coefficient to express the rate of the reaction.

Rate of reaction $= - \frac{d [A]}{d t} = - \frac{1}{2} \frac{d [B]}{d t}$