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Hint: As volume of any gas gets changed, it also affects its concentration. Rate of reaction and velocity of reaction are both the same. The formula of rate of reaction of this reaction is

$${\text{Rate of reaction = k}}{\left[ {S{O_{2(g)}}} \right]^2}\left[ {{O_{2(g)}}} \right]$$.

Complete Step-by-Step Solution:

Here, two different conditions are given in the question in which reaction is carried out in vessels having different volumes. We know that concentration of any gaseous species depends on the volume of the vessel as well. So, as this reaction is happening in two vessels having different volumes, the concentration of the products and reactants will not be the same.

Let’s find out the relation between velocities of reaction in two different vessels.

We know that for given reaction,

$${\text{Rate of reaction = k}}{\left[ {S{O_{2(g)}}} \right]^2}\left[ {{O_{2(g)}}} \right]$$

Now, if we consider concentration of $$S{O_{2(g)}}$$ in 1$$d{m^3}$$ vessel as x, then concentration of $$S{O_{2(g)}}$$ in 2 $$d{m^3}$$ vessel will be $$\dfrac{x}{2}$$ because as volume of any gas gets doubled, the concentration gets halved.

Same way, suppose concentration of $${O_{2(g)}}$$ in 1$$d{m^3}$$ vessel is equal to y, then concentration of $${O_{2(g)}}$$ in 2 $$d{m^3}$$ vessel will be $$\dfrac{y}{2}$$. Now putting these concentration values in the rate of reaction equation, we get two new equations as follows. Suppose vessel having 1$$d{m^3}$$ volume as vessel-1 and other vessel as vessel-2

For Vessel-1, $${\text{Rate of reaction = k}}{\left[ x \right]^2}\left[ y \right]$$……………..(1)

For Vessel-2, $${\text{Rate of reaction = k}}{\left[ {\dfrac{x}{2}} \right]^2}\left[ {\dfrac{y}{2}} \right]$$……………….(2)

Dividing equation (1) and (2), we get

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{{{\text{k}}{{\left[ x \right]}^2}\left[ y \right]}}{{{\text{k}}{{\left[ {\dfrac{x}{2}} \right]}^2}\left[ {\dfrac{y}{2}} \right]}}$$

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{{{x^2}y}}{{\dfrac{{{x^2}y}}{8}}}$$

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{8}{1}$$

Additional Information:

-When these types of ratios of rate of reactions are being asked, we can simply put supposed values of concentration into the equation of rate of reaction and have the correct ratio.

- Reaction rate is a measure of increase in concentration of products or decrease in concentration of reactants per unit time.

Note: Make sure that numerator and denominator are respectively true in case of the questions in which the ratio is asked. Do not forget that there are two molecules of $$S{O_{2(g)}}$$ that are involved in a single reaction, so the rate of reaction will have an effect on that number.

$${\text{Rate of reaction = k}}{\left[ {S{O_{2(g)}}} \right]^2}\left[ {{O_{2(g)}}} \right]$$.

Complete Step-by-Step Solution:

Here, two different conditions are given in the question in which reaction is carried out in vessels having different volumes. We know that concentration of any gaseous species depends on the volume of the vessel as well. So, as this reaction is happening in two vessels having different volumes, the concentration of the products and reactants will not be the same.

Let’s find out the relation between velocities of reaction in two different vessels.

We know that for given reaction,

$${\text{Rate of reaction = k}}{\left[ {S{O_{2(g)}}} \right]^2}\left[ {{O_{2(g)}}} \right]$$

Now, if we consider concentration of $$S{O_{2(g)}}$$ in 1$$d{m^3}$$ vessel as x, then concentration of $$S{O_{2(g)}}$$ in 2 $$d{m^3}$$ vessel will be $$\dfrac{x}{2}$$ because as volume of any gas gets doubled, the concentration gets halved.

Same way, suppose concentration of $${O_{2(g)}}$$ in 1$$d{m^3}$$ vessel is equal to y, then concentration of $${O_{2(g)}}$$ in 2 $$d{m^3}$$ vessel will be $$\dfrac{y}{2}$$. Now putting these concentration values in the rate of reaction equation, we get two new equations as follows. Suppose vessel having 1$$d{m^3}$$ volume as vessel-1 and other vessel as vessel-2

For Vessel-1, $${\text{Rate of reaction = k}}{\left[ x \right]^2}\left[ y \right]$$……………..(1)

For Vessel-2, $${\text{Rate of reaction = k}}{\left[ {\dfrac{x}{2}} \right]^2}\left[ {\dfrac{y}{2}} \right]$$……………….(2)

Dividing equation (1) and (2), we get

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{{{\text{k}}{{\left[ x \right]}^2}\left[ y \right]}}{{{\text{k}}{{\left[ {\dfrac{x}{2}} \right]}^2}\left[ {\dfrac{y}{2}} \right]}}$$

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{{{x^2}y}}{{\dfrac{{{x^2}y}}{8}}}$$

$$\dfrac{{{\text{Rate of reaction of vessel - 1}}}}{{{\text{Rate of reaction of vessel - 2}}}} = \dfrac{8}{1}$$

__So, correct answer is (D) 8 : 1__Additional Information:

-When these types of ratios of rate of reactions are being asked, we can simply put supposed values of concentration into the equation of rate of reaction and have the correct ratio.

- Reaction rate is a measure of increase in concentration of products or decrease in concentration of reactants per unit time.

Note: Make sure that numerator and denominator are respectively true in case of the questions in which the ratio is asked. Do not forget that there are two molecules of $$S{O_{2(g)}}$$ that are involved in a single reaction, so the rate of reaction will have an effect on that number.

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