
A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will______.
A.Be tripled
B.Increase by a factor $\,4\,$
C.Be doubled
D.Remain unchanged
Answer
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Hint: The reaction mentioned in the question is of second order kinetics even though the only reactant is carbon monoxide. So, in order to find the rate of the reaction, we have to use the rate law equation. Then, we can find the change in rate when the concentration of carbon monoxide is doubled.
Complete step by step answer:
Let us first understand the rate law equation;
The rate law for a chemical reaction is an equation that applies the reaction rate to the reactants' concentrations or partial pressures.
So, for a general reaction $\,aA + bB \to C\,$, the rate law can be stated as;
$\,r = k{[A]^x}{[B]^y}\,$
In this equation, $\,[A]\,$ and $\,[B]\,$ express concentrations of the reactants in units of $\,moles/litre\,$. For each reaction, the exponents $\,x\,$ and $\,y\,$ are different and must be experimentally determined; they are not related to the chemical equation's stoichiometric coefficients. Finally, the rate constant of the response is defined as $\,k\,$ . The value of this k coefficient can differ according to the factors influencing the rate of reaction, such as pressure, temperature, surface area, etc. A smaller rate constant implies a slower reaction, while a quicker reaction is indicated by a larger rate constant.
Now, coming to the question, we can write the rate law expression as;
$\,R = k{[CO]^2}\,$
So, if the concentration of carbon monoxide is doubled, then the rate reaction will be;
$\,R' = k{[2CO]^2}\,$
$\, \Rightarrow 4k{[CO]^2} = 4R\,$
Hence, we understand that as the concentration of carbon monoxide is doubled, the rate law is increased by a factor of $\,4\,$.
So, for this question, option A is the correct answer.
Note:
In general, it is not possible to deduce the rate equation of a reaction with a multi-step process from the stoichiometric coefficients of the total reaction; it must be calculated experimentally. For elementary reactions, using collision theory, the rate equation may be constructed from first principles.
Complete step by step answer:
Let us first understand the rate law equation;
The rate law for a chemical reaction is an equation that applies the reaction rate to the reactants' concentrations or partial pressures.
So, for a general reaction $\,aA + bB \to C\,$, the rate law can be stated as;
$\,r = k{[A]^x}{[B]^y}\,$
In this equation, $\,[A]\,$ and $\,[B]\,$ express concentrations of the reactants in units of $\,moles/litre\,$. For each reaction, the exponents $\,x\,$ and $\,y\,$ are different and must be experimentally determined; they are not related to the chemical equation's stoichiometric coefficients. Finally, the rate constant of the response is defined as $\,k\,$ . The value of this k coefficient can differ according to the factors influencing the rate of reaction, such as pressure, temperature, surface area, etc. A smaller rate constant implies a slower reaction, while a quicker reaction is indicated by a larger rate constant.
Now, coming to the question, we can write the rate law expression as;
$\,R = k{[CO]^2}\,$
So, if the concentration of carbon monoxide is doubled, then the rate reaction will be;
$\,R' = k{[2CO]^2}\,$
$\, \Rightarrow 4k{[CO]^2} = 4R\,$
Hence, we understand that as the concentration of carbon monoxide is doubled, the rate law is increased by a factor of $\,4\,$.
So, for this question, option A is the correct answer.
Note:
In general, it is not possible to deduce the rate equation of a reaction with a multi-step process from the stoichiometric coefficients of the total reaction; it must be calculated experimentally. For elementary reactions, using collision theory, the rate equation may be constructed from first principles.
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