Answer :
Answer : The order of reaction with respect to A is, first order reaction.
The order of reaction with respect to B is, zero order reaction.
The overall order of reaction is, first order reaction.
Explanation :
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
For the given chemical equation:
[tex]A+B\rightarrow Products[/tex]
Rate law expression for the reaction:
[tex]\text{Rate}=k[A]^a[B]^b[/tex]
where,
a = order with respect to A
b = order with respect to B
Expression for rate law for first observation:
[tex]3.20\times 10^{-1}=k(1.50)^a(1.50)^b[/tex] ....(1)
Expression for rate law for second observation:
[tex]3.20\times 10^{-1}=k(1.50)^a(2.50)^b[/tex] ....(2)
Expression for rate law for third observation:
[tex]6.40\times 10^{-1}=k(3.00)^a(1.50)^b[/tex] ....(3)
Dividing 1 from 2, we get:
[tex]\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0[/tex]
Dividing 1 from 3, we get:
[tex]\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1[/tex]
Thus, the rate law becomes:
[tex]\text{Rate}=k[A]^1[B]^0[/tex]
[tex]\text{Rate}=k[A][/tex]
Thus,
The order of reaction with respect to A is, first order reaction.
The order of reaction with respect to B is, zero order reaction.
The overall order of reaction is, first order reaction.
The order of reaction with respect to A is 1
Consider the given reaction:
A + B → Products
Using the rate equation to determine the order of reaction, we have:
[tex]\mathbf{rate = K[A]^x [B]^y}[/tex]
From the given table in the question, if the rate in m/s for the first term is:
3.20 × 10⁻¹
Then, we have:
[tex]\mathbf{3.20 \times 10^{-1} = k [1.5]^x [1.5]^y ---- (1)}[/tex]
For the second equation, we have:
[tex]\mathbf{3.20 \times 10^{-1} = k [1.5]^x [2.5]^y ---- (2)}[/tex]
The third equation, we have:
[tex]\mathbf{6.40 \times 10^{-1} = k [1.5]^x [3.0]^y ---- (3)}[/tex]
Now, if we equate equations (1) and (3), we have:
[tex]\mathbf{\dfrac{6.40 \times 10^{-1}}{3.2 \times 10^{-1}} = \dfrac{[3.0]^x [1.5]^y}{[1.5]^x[1.5]^y} }[/tex]
[tex]\mathbf{2^1 = \Big(\dfrac{3.0}{1.5} \Big) ^x}[/tex]
[tex]\mathbf{2^1 =2 ^x}[/tex]
∴
x = 1
From equation (1) and (2);
[tex]\mathbf{\dfrac{3.20 \times 10^{-1}}{3.20 \times 10^{-1}} = \dfrac{[1.5]^x [2.5]^y}{[1.5]^x[1.5]^y} }[/tex]
[tex]\mathbf{1 = 1.66 ^y}[/tex]
[tex]\mathbf{y = 0}[/tex]
Hence, the rate of the reaction with regard to A = 1 and the rate of reaction with respect to B = 0.
Therefore, we can conclude that the order of the reaction is one 1+ 0 = 1
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