
for the irreversible process, A+B→Products, the rate is first order w.r.t. A and second order w.r.t. B. If one mole each of “A and B” are introduced into a 1.0L vessel, and the initial rate was $1.0 \times {10^{ - 2}}mole/lt/s$ when half of the reactants have been turned into products are ___?
A) $1.25 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
B) $1 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
C) $2.5 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
D) $2 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
Answer
569.4k+ views
Hint: The mathematical expression showing the dependence of rate on the concentration(s) of reactant(s) is known as rate-law or rate-expression of the reaction and sum of the indices (powers) of the concentration terms appearing in the rate law as observed experimentally is called order of reaction
Complete Step by step answer: In general, if rate law of a reaction represented by the equation.
\[aA{\text{ }} + {\text{ }}bB \to \]Products
is experimentally found to be as follows:
\[Rate \propto {\left[ A \right]^m}{\left[ B \right]^n}\]
Then order w.r.t. \[A = m\]
order w.r.t. \[B = n\]
overall order \[ = m + n\]
It is to be noted that ‘m’ may or may not be equal to ‘a’ and similarly. ‘n’ may or may not be equal to ‘b’. m and n are experimental quantities and their values which really depend on the reaction-mechanism and experimental condition, may not be predicted by just seeing the chemical equation of the reaction.
The Reaction is 1st order wrt A and 2nd order wrt B.
Initial number of moles of A and B are 1 mol and 1 mol respectively
So, $Rate = K{[A]^1}{[B]^2}$
When only half of reactants are converted and half are unreacted
\[\left[ A \right] = 0.5{\text{ }}mol{\text{ }};{\text{ }}\left[ B \right] = 0.5{\text{ }}mol\]
$Rate = [0.5 \times A]{[0.5 \times B]^2}$ --(1)
$Rate\; = 0.125\,mol{L^{ - 1}}{S^{ - 1}} = 1.25 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
Hence Option “A” is correct.
Note: Reactions with some kind of chemical equations may differ in this rate laws and hence order. Number of molecules of the reactant(s) whose concentration changes during the chemical change is called order of reaction. Equation (1) comes from the first that half the reaction is completed.
Complete Step by step answer: In general, if rate law of a reaction represented by the equation.
\[aA{\text{ }} + {\text{ }}bB \to \]Products
is experimentally found to be as follows:
\[Rate \propto {\left[ A \right]^m}{\left[ B \right]^n}\]
Then order w.r.t. \[A = m\]
order w.r.t. \[B = n\]
overall order \[ = m + n\]
It is to be noted that ‘m’ may or may not be equal to ‘a’ and similarly. ‘n’ may or may not be equal to ‘b’. m and n are experimental quantities and their values which really depend on the reaction-mechanism and experimental condition, may not be predicted by just seeing the chemical equation of the reaction.
The Reaction is 1st order wrt A and 2nd order wrt B.
Initial number of moles of A and B are 1 mol and 1 mol respectively
So, $Rate = K{[A]^1}{[B]^2}$
When only half of reactants are converted and half are unreacted
\[\left[ A \right] = 0.5{\text{ }}mol{\text{ }};{\text{ }}\left[ B \right] = 0.5{\text{ }}mol\]
$Rate = [0.5 \times A]{[0.5 \times B]^2}$ --(1)
$Rate\; = 0.125\,mol{L^{ - 1}}{S^{ - 1}} = 1.25 \times {10^{ - 3}}mol{L^{ - 1}}{S^{ - 1}}$
Hence Option “A” is correct.
Note: Reactions with some kind of chemical equations may differ in this rate laws and hence order. Number of molecules of the reactant(s) whose concentration changes during the chemical change is called order of reaction. Equation (1) comes from the first that half the reaction is completed.
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