Question

In the reaction of the formation of sulphur trioxide by contact process; $\text{ 2S}{{\text{O}}_{\text{2}}}\text{+}{{\text{O}}_{\text{2}}}\rightleftharpoons \text{2S}{{\text{O}}_{\text{3}}}\text{ }$, the rate of reaction was measured as $\text{ }\dfrac{\text{d}\left[ {{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{=-2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\text{ }$ . The rate of disappearance of $\left[ \text{S}{{\text{O}}_{\text{2}}} \right]$ in $\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}\text{ }$ will be:
(A) $\text{ }-3.75\times {{10}^{-4}}$
(B) $\text{ }-1.25\times {{10}^{-4}}$
(C) $\text{ }-2.50\times {{10}^{-4}}$
(D) $\text{ }-5.00\times {{10}^{-4}}$

Answer 1

Hint: We know that according to kinetics, the rate of a reaction is equal to the rate of disappearance of the reactants or the rate of formation of products.
If a reaction is$\text{ A + B }\rightleftharpoons \text{ C + D }$,
Then the rate of this reaction is given as,
$\text{ }\dfrac{-\text{d}\left[ \text{A} \right]}{\text{dt}}\text{=}\dfrac{-\text{d}\left[ \text{B} \right]}{\text{dt}}\text{=}\dfrac{\text{d}\left[ \text{C} \right]}{\text{dt}}\text{=}\dfrac{\text{d}\left[ \text{D} \right]}{\text{dt}}\text{ }$.
Where, the $\left[ \text{A} \right]$ ,$\left[ \text{B} \right]$ , $\left[ \text{C} \right]$ and $\left[ \text{D} \right]$ are the concentration of reactant and product.

Complete step by step solution:
Chemical Kinetics is a branch of chemistry that deals with the study of the rate of a reaction.
We know that the rate of reaction is the rate of disappearance of reactants or the rate of formation of products.
For a reaction, the rate of reaction can be written in terms of the concentration of the product or reactant. For a balanced chemical reaction, the stoichiometric coefficients of the reactant and product are written in the denominator with their rate expression.
We are given with a reaction of $\text{S}{{\text{O}}_{\text{2}}}$ with${{\text{O}}_{\text{2}}}$ that produces$\text{S}{{\text{O}}_{\text{3}}}$.
The reaction is as follows:
$\text{ 2S}{{\text{O}}_{\text{2}}}\text{+}{{\text{O}}_{\text{2}}}\rightleftharpoons \text{2S}{{\text{O}}_{\text{3}}}\text{ }$

From the reaction,
The rate of disappearance of reactants i.e. $\text{S}{{\text{O}}_{\text{2}}}$ would be equal to $\dfrac{-1}{\text{ }2}\dfrac{\text{d}\left[ \text{S}{{\text{O}}_{\text{2}}} \right]}{\text{dt}}$ . Since the 2 molecules of $\text{ S}{{\text{O}}_{\text{2}}}\text{ }$are taking part in the reaction.
We know that the rate of reaction of the reactant is equal. Thus, the rate of disappearance of $\text{S}{{\text{O}}_{\text{2}}}$ will is equal to the disappearance of ${{\text{O}}_{\text{2}}}$ and equal to the formation of$\text{S}{{\text{O}}_{\text{3}}}$.
Therefore, the relation can be given as,
$\dfrac{\text{d}\left[ {{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{ = }\dfrac{1}{2}\dfrac{\text{d}\left[ \text{S}{{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{= }-\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{mol}{{\text{L}}^{\text{-1}}}{{\text{s}}^{\text{-1}}}$
Since, we are given that,
$\text{Rate of reaction =}~-\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{mol}{{\text{L}}^{\text{-1}}}{{\text{s}}^{\text{-1}}}$
So, the rate of disappearance of $\left[ \text{S}{{\text{O}}_{\text{2}}} \right]$ will be,
$\begin{align}
& \text{ }\dfrac{1}{2}\dfrac{\text{d}\left[ \text{S}{{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{= }-\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}} \\
& \text{ }\dfrac{\text{d}\left[ \text{S}{{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{= 2}\times \text{(}-\text{2}\text{.5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}) \\
& \therefore \dfrac{\text{ d}\left[ \text{S}{{\text{O}}_{\text{2}}} \right]}{\text{dt}}\text{= }-5.0\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}} \\
\end{align}$
Therefore, the rate of disappearance of $\text{ }\left[ \text{S}{{\text{O}}_{\text{2}}} \right]\text{ }$will be$-5.0\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-4}}}\text{ mol }{{\text{L}}^{\text{-1}}}\text{ }{{\text{s}}^{\text{-1}}}$.

Hence, (D) is correct.

Note: Remember that the rate of reaction is always a non-negative value. It can be zero or not but it must be an integer. The negative sign before the concentration of reactant is because the concentration of reactant decreases and for a product the concentration goes on the increase, thus it has a positive sign. Always start with a balanced chemical reaction.