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Question: In the following reaction; \(xA \to yB\) \({\log _{10}}\left[ { - \dfrac{{d\left[ A \right]}}{{dt}...

In the following reaction; xA→yBxA \to yB
log⁡10[−d[A]dt]=log⁡10[d[B]dt]+0.3010{\log _{10}}\left[ { - \dfrac{{d\left[ A \right]}}{{dt}}} \right] = {\log _{10}}\left[ {\dfrac{{d\left[ B \right]}}{{dt}}} \right] + 0.3010
A and B respectively can be:
A.n-Butane and iso-butane.
B.C22H4&C4H8{{\text{C}}_2{2}}{{\text{H}}_4}\& {{\text{C}}_4}{{\text{H}}_8}
C.N2O4&NO2{{\text{N}}_2}{{\text{O}}_4}\& {\text{N}}{{\text{O}}_2}
D.C2H2&C6H6{{\text{C}}_2}{{\text{H}}_2}\& {{\text{C}}_6}{{\text{H}}_6}

Explanation

Solution

We know that the Reaction Rate determines the speed of a chemical reaction and it is defined as the change in concentration of the reactants or the change in concentration of the products per unit time.
For the reactionA→BA \to B. The rate of the reaction is given by,
rate=dAdtrate = \dfrac{{dA}}{{dt}}

Complete step by step answer:
Given,
The rate law of the equation is,
log⁡10[−d[A]dt]=log⁡10[d[B]dt]+0.3010{\log _{10}}\left[ { - \dfrac{{d\left[ A \right]}}{{dt}}} \right] = {\log _{10}}\left[ {\dfrac{{d\left[ B \right]}}{{dt}}} \right] + 0.3010
We know that the value of log⁡2=0.3010\log 2 = 0.3010.
CH3COCH2CH2COOC2H5C{H_3}COC{H_2}C{H_2}COO{C_2}{H_5}
The minus sign in the reaction rate indicates the decrease in concentration of the reactant while the positive sign indicates the increase in the concentration of the product.
Taking log on both sides,
[−d[A]dt]=2[d[B]dt]\left[ { - \dfrac{{d\left[ A \right]}}{{dt}}} \right] = 2\left[ {\dfrac{{d\left[ B \right]}}{{dt}}} \right]
12[−d[A]dt]=[d[B]dt]\dfrac{1}{2}\left[ { - \dfrac{{d\left[ A \right]}}{{dt}}} \right] = \left[ {\dfrac{{d\left[ B \right]}}{{dt}}} \right]
From the rate law, the reaction is determined as 2A→B2A \to B. Thus option B is correct.
The reaction is,
2C2H4→C4H82{C_2}{H_4}\xrightarrow{{}}{C_4}{H_8}
Therefore, the option B is correct.

Note: The equation that relates that rate of a reaction to the rate constant and the concentrations of the reactants are called rate law; K is the rate constant for a given reaction. The general rate law is usually expressed as:
Rate=k[A]s[B]tRate = k{\left[ A \right]^s}{\left[ B \right]^t}
Reaction order:
The reaction rate for a given reaction is used to calculate the order of a reaction. The order of a reaction is significantly used to arrange chemical reactions easily. Reaction order is calculated from the rate law by adding the exponential values of the reactants in the rate law.
order=s+torder = s + t
It is significant to note that even though the reaction order is determined from the rate law, there is no relationship between the reaction order and the stoichiometric coefficients in the chemical equation.