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Question: In a reaction, \(2{{X}} \to {{Y}}\), the concentration of \({{X}}\) decreases from \(0.50{{M}}\) to ...

In a reaction, 2X→Y2{{X}} \to {{Y}}, the concentration of X{{X}} decreases from 0.50M0.50{{M}} to 0.38M0.38{{M}} in 10min{{10min}}. What is the rate of reaction in Ms−1{{M}}{{{s}}^{ - 1}} during this interval?
A. 2×10−42 \times {10^{ - 4}}
B. 4×10−24 \times {10^{ - 2}}
C. 2×10−22 \times {10^{ - 2}}
D. 1×10−21 \times {10^{ - 2}}

Explanation

Solution

The part of chemical science which deals with the study of the rates of chemical reactions and the factors which affect the rate of reactions is called the chemical kinetics. Rate of a chemical reaction can be described as the change in concentration of a reactant or product with time.

Complete step by step solution:
The change in any variable in unit time is termed as the rate. Rates can be determined by focusing on the concentration as a function of time. When the reaction proceeds, the amount of reactants decreases and the product increases. Thus the rate of products is positive and rate of reactants is negative.
Consider a reaction A→B{{A}} \to {{B}}
We can express rate of this reaction as r=−1aΔ[A]Δt=1bΔ[B]Δt{{r}} = - \dfrac{1}{{{a}}}\dfrac{{\Delta \left[ {{A}} \right]}}{{\Delta {{t}}}} = \dfrac{1}{{{b}}}\dfrac{{\Delta \left[ {{B}} \right]}}{{\Delta {{t}}}}, where a,b{{a,b}} are the stoichiometric coefficients of A{{A}} and B{{B}} respectively.
Also, change in reactant concentration, ΔA=[A]f−[A]i\Delta {{A}} = {\left[ {{A}} \right]_{{f}}} - {\left[ {{A}} \right]_{{i}}}
Now we can consider the reaction given in the question.
i.e. 2X→Y2{{X}} \to {{Y}}
When the rate can be expressed as just the change in concentration in unit time, the calculation will be simpler, i.e. r=−Δ[X]Δt=Δ[Y]Δt{{r}} = - \dfrac{{\Delta \left[ {{X}} \right]}}{{\Delta {{t}}}} = \dfrac{{\Delta \left[ {{Y}} \right]}}{{\Delta {{t}}}}
Δ[X]=0.50−0.38=0.12molL−1\Delta \left[ {{X}} \right] = 0.50 - 0.38 = 0.12{{mol}}{{{L}}^{ - 1}} and the time, t=10min=10×60s=600s{{t = 10min = 10}} \times {{60s = 600s}}
We can consider the rate with respect to the reactants since the concentration of reactant is given.
[X]f=0.50molL−1{\left[ {{X}} \right]_{{f}}} = 0.50{{mol}}{{{L}}^{ - 1}}, [X]i=0.38molL−1{\left[ {{X}} \right]_{{i}}} = 0.38{{mol}}{{{L}}^{ - 1}}
Substitute the value of Δ[X]\Delta \left[ {{X}} \right] in the rate equation, we get
r=−0.12molL−1600s=2×10−4Ms−1{{r}} = - \dfrac{{0.12{{mol}}{{{L}}^{ - 1}}}}{{600{{s}}}} = 2 \times {10^{ - 4}}{{M}}{{{s}}^{ - 1}}
So the rate of reaction is 2×10−4Ms−12 \times {10^{ - 4}}{{M}}{{{s}}^{ - 1}}.

Thus, the correct option will be A.

Note: Generally rate of a reaction is determined using an equation with respect to stoichiometric coefficients.
i.e. r=−1aΔ[X]Δt=1bΔ[Y]Δt{{r}} = - \dfrac{1}{{{a}}}\dfrac{{\Delta \left[ {{X}} \right]}}{{\Delta {{t}}}} = \dfrac{1}{{{b}}}\dfrac{{\Delta \left[ {{Y}} \right]}}{{\Delta {{t}}}}
In this reaction, stoichiometric coefficient of X{{X}} is 22 and Y{{Y}} is 11, i.e. a=2{{a = 2}} and b=1{{b = 1}}.
Therefore rate can be represented as r=−12Δ[X]Δt=11Δ[Y]Δt{{r}} = - \dfrac{1}{2}\dfrac{{\Delta \left[ {{X}} \right]}}{{\Delta {{t}}}} = \dfrac{1}{1}\dfrac{{\Delta \left[ {{Y}} \right]}}{{\Delta {{t}}}}
r=−12×0.12molL−1600s=1×10−5molL−1s=1×10−5Ms−1{{r}} = - \dfrac{1}{2} \times \dfrac{{0.12{{mol}}{{{L}}^{ - 1}}}}{{600{{s}}}} = 1 \times {10^{ - 5}}{{mol}}{{{L}}^{ - 1}}{{s = }}1 \times {10^{ - 5}}{{M}}{{{s}}^{ - 1}}
Thus we can say that the rate of reaction is 1×10−5Ms−11 \times {10^{ - 5}}{{M}}{{{s}}^{ - 1}}.