Question:
During the kinetic study of the reaction, $2A + B\longrightarrow C + D$, following results were obtained . Based on the above data which one of the following is correct?
Run
$A/mol\,L^{-1}$
$B/mol\,L^{-1}$
Initial rate of formation of $D/mol\,L^{-1}\,min^{-1}$
I
0.1
0.1
$6.0 \times 10^{-3}$
II
0.3
0.2
$7.2 \times 10^{-2}$
III
0.3
0.4
$2.88 \times 10^{-1}$
IV
0.4
0.1
$2.40 \times 10^{-2}$
During the kinetic study of the reaction, $2A + B\longrightarrow C + D$, following results were obtained . Based on the above data which one of the following is correct?
| Run | $A/mol\,L^{-1}$ | $B/mol\,L^{-1}$ | Initial rate of formation of $D/mol\,L^{-1}\,min^{-1}$ |
|---|---|---|---|
| I | 0.1 | 0.1 | $6.0 \times 10^{-3}$ |
| II | 0.3 | 0.2 | $7.2 \times 10^{-2}$ |
| III | 0.3 | 0.4 | $2.88 \times 10^{-1}$ |
| IV | 0.4 | 0.1 | $2.40 \times 10^{-2}$ |
Updated On: May 5, 2024
- Rate = k$[A]^2$[B]
- Rate = k[A][B]
- Rate = k$[A]^2 [B]^2$
- Rate = k[A]$[B]^2$
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The Correct Option is D
Solution and Explanation
Let the order of reaction with respect to A is x
and with respect to B is y. Thus,
rate=k$[A]^x [B]^y$
($x$ and $y$ are stoichiometric coefficient)
For the given cases,
I. rate = $k (0.1)^x \, (0.1)^y =6.0 \times 10^{-3}$
II. rate = $k (0.3)^x \, (0.2)^y =7.2 \times 10^{-2}$
III. rate = $k (0.3)^x \, (0.40)^y =2.88 \times 10^{-1}$
IV. rate = $k (0.34)^x \, (0.1)^y =2.40 \times 10^{-2}$
Dividing E (I) by E (IV), we get
$ \big(\frac{0.1}{0.4}\big)^x \big(\frac{0.1}{0.1}\big)^y =\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$
or $ \big(\frac{1}{4} \big)^x = \big(\frac{1}{4} \big)^1$
$\therefore \hspace25mm x=1$
On dividing E (II) by E (Ill), we get
$ \big(\frac{0.3}{0.3}\big)^x \big(\frac{0.2}{0.4}\big)^y =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$
or $\big(\frac{1}{2} \big)^y = \frac{1}{4}$
or $ \big(\frac{1}{2} \big)^y = \big(\frac{1}{2} \big)^2$
$\therefore \hspace25mm y=2$
Thus, rate law is,
$ rate = k[A]^1 [B]^2$
$ = k[A][B]^2$
and with respect to B is y. Thus,
rate=k$[A]^x [B]^y$
($x$ and $y$ are stoichiometric coefficient)
For the given cases,
I. rate = $k (0.1)^x \, (0.1)^y =6.0 \times 10^{-3}$
II. rate = $k (0.3)^x \, (0.2)^y =7.2 \times 10^{-2}$
III. rate = $k (0.3)^x \, (0.40)^y =2.88 \times 10^{-1}$
IV. rate = $k (0.34)^x \, (0.1)^y =2.40 \times 10^{-2}$
Dividing E (I) by E (IV), we get
$ \big(\frac{0.1}{0.4}\big)^x \big(\frac{0.1}{0.1}\big)^y =\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}$
or $ \big(\frac{1}{4} \big)^x = \big(\frac{1}{4} \big)^1$
$\therefore \hspace25mm x=1$
On dividing E (II) by E (Ill), we get
$ \big(\frac{0.3}{0.3}\big)^x \big(\frac{0.2}{0.4}\big)^y =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}$
or $\big(\frac{1}{2} \big)^y = \frac{1}{4}$
or $ \big(\frac{1}{2} \big)^y = \big(\frac{1}{2} \big)^2$
$\therefore \hspace25mm y=2$
Thus, rate law is,
$ rate = k[A]^1 [B]^2$
$ = k[A][B]^2$
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Factors Affecting The Reaction Rate:
- The concentration of Reactants - According to collision theory, which is discussed later, reactant molecules collide with each other to form products.
- Nature of the Reactants - The reaction rate also depends on the types of substances that are reacting.
- Physical State of Reactants - The physical state of a reactant whether it is solid, liquid, or gas can greatly affect the rate of change.
- Surface Area of Reactants - When two or more reactants are in the same phase of fluid, their particles collide more often than when either or both are in the solid phase or when they are in a heterogeneous mixture. In a heterogeneous medium, the collision between the particles occurs at an interface between phases. Compared to the homogeneous case, the number of collisions between reactants per unit time is significantly reduced, and so is the reaction rate.
- Temperature - If the temperature is increased, the number of collisions between reactant molecules per second. Increases, thereby increasing the rate of the reaction.
- Effect Of Solvent - The nature of the solvent also depends on the reaction rate of the solute particles.
- Catalyst - Catalysts alter the rate of the reaction by changing the reaction mechanism.