Question

In the presence of vanadium oxide, SO₂(g) reacts with an excess of oxygen to give SO₃(g):

\(S{O_2}\; + \;\dfrac{1}{2}{O_2}\; \to \;S{O_3}\)

This reaction is an important step in the manufacture of sulfuric acid. It is observed that tripling the SO₂ concentration increases the rate by a factor of 3, but tripling the SO₃concentration decreases the rate by a factor of\(1.7 \approx \;\sqrt 3 \). The rate is insensitive to the O₂ concentration as long as an excess of oxygen is present

(a) Write the rate expression for this reaction, and give the units of the rate constant k.

(b) If (SO₂) is multiplied by 2 and (SO₃) by 4 but all other conditions are unchanged, what change in the rate will be observed?

Short answer

The new reaction rate increases by four times

R = 2.4R

Step-by-step solution

STEP-1: Rate expression for a reaction

a.The rate of chemical reaction can be explained as expression which describes relation between rate and product of concendration of reactents raised to power.

Rate expression for chemical reaction of the reactents A and B is given by expression

\(Rate\; = \;K{\left( A \right)^m}{\left( B \right)^n}\)

Where k is the rate constant of reaction m and n are the order of the reaction

With respect to species A and species B.

STEP-2: Rate expression:

For the given chemical reaction

\(S{O_2}\; + \;\dfrac{1}{2}{O_2}\; \to \;S{O_3}\)

The rate of chemical reaction depends on the concendration of \(S{O_2}\;\)andSO_3.The expression can be written as

\(R\; = \;k{\left( {S{O_2}} \right)^m}\;{\left( {S{O_3}} \right)^n}............\left( 1 \right)\)

The expression for the rate law when the concendration of SO₂ is increased by two times and concendration of SO₃ is increased by 4 times then the expression is written as and also it was found that rate increases by 3 times finally

\(3R\; = \;k\;{\left( {3S{O_2}} \right)^m}{\left( {S{O_3}} \right)^n}.................\left( 2 \right)\)

Expression for the rate law when the rate is decreased by the factor of\(\sqrt 3 \)and on tripling the concendration of \(S{O_3}\)is as follows

\(\sqrt 3 \; = \;k{\left( {S{O_2}} \right)^m}{\left( {3S{O_3}} \right)^n}.........\left( 3 \right)\)

STEP-3:  Order  of reaction :

Divide the equation 2 by 1 to obtain the values of m

\(\begin{aligned}{l}\dfrac{{3R}}{R}\; = \;\;\dfrac{{k{{\left( {3S{O_2}} \right)}^m}{{\left( {S{O_3}} \right)}^n}}}{{k{{\left( {S{O_2}} \right)}^m}{{\left( {S{O_3}} \right)}^n}}}........\left( 4 \right)\\3\; = \;{3^m}\\{\left( 3 \right)^1}\; = \;{3^m}\\m\; = \;1\end{aligned}\)

Similarly divide the equation 3 by 1 to get the values of n\(\begin{aligned}{l}\dfrac{{\sqrt 3 R}}{R}\; = \;\;\dfrac{{k\left( {S{O_2}} \right){{\left( {3S{O_3}} \right)}^n}}}{{k\left( {S{O_2}} \right){{\left( {S{O_3}} \right)}^n}}}........\left( 5 \right)\\\sqrt 3 \; = \;{3^n}\\{\left( 3 \right)^{\dfrac{1}{2}}}\; = \;{3^n}\\n\; = \;\dfrac{1}{2}\end{aligned}\)

The overall order of the reaction can be written as

\(\begin{aligned}{l}Overallorder\\\;x\; = \;m\; + \;n\\x\;\; = \;1\; + \;\dfrac{1}{2}\\x\; = \;\dfrac{3}{2}\end{aligned}\)

Substitute 1 for m and\(n\; = \dfrac{1}{2}\)rate law expression is givn as

\(Rate\; = \;k\left( {S{O_2}} \right){\left( {S{O_3}} \right)^{\dfrac{1}{2}}}.........\left( 6 \right)\)

Units of rate constant by substituting m and n values following expression is derived