Q.

In the reaction 2SO2(g) + O2(g) 2SO3(g), 2 moles of SO2 ,1 mole of O2 and 2 moles of SO3 are present in equilibrium. what is the number of moles of O2 be introduced into the vessel to increase the equilibrium moles of SO3 to 3 moles

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a

9

b

8

c

7.5

d

8.5

answer is D.

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Detailed Solution

Following is the given equilibrium,

\large \begin{array}{*{20}{c}} {\mathop {}\limits_{t = {t_{eq}}} } \\ {} \end{array}\begin{array}{*{20}{c}} {{{\mathop {2S{O_2}\left( g \right)}\limits_2 }_,} + \mathop {{O_2}\left( g \right)}\limits_1 \rightleftharpoons \mathop {2S{O_3}\left( g \right)}\limits_2 } \\ {} \end{array}\,

The volume of the container is not given let us assume it to be 1 lit

Let the moles of O2 to be introduced so that the moles of SO3 becomes 3 at equilibrium be x.

Again,

                    

 2SO2+O2
\large \rightleftharpoons
2SO3(g)
t = teq2-2y 
\large 1 + x - y
 2 +2 y

Now, 2 +2 y = 3 ⇒ y = 12   and at same temperature KC remains constant, equating both Kvalues

\large {K_C} = \frac{{{{\left( 3 \right)}^2}}}{{\left( {\frac{1}{2} + x} \right){{\left( 1 \right)}^2}}} = \frac{{{2^2}}}{{1 \times {2^2}}} = 1
\large \Rightarrow 9 = \frac{1}{2} + x \Rightarrow x = 8.5
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