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Q.

In a 500 ml flask, the degree of dissociation of PCl5 at equilibrium is 40% when the initial amount taken  is 5 moles. The value of equilibrium constant in moles/lit  for the decomposition of PCl5 is ( Given reaction, PCl5gPCl3g+Cl2g )

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a

3.33

b

2.66

c

5.32

d

4.66

answer is B.

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Detailed Solution

Vvessel = 0.5 lit

 

Degree of decomposition of PCl5 =

\large \frac{{40}}{{100}} = 0.4
\large {\left( {{n_{PC{l_5}}}} \right)_i} = 5

.

For every 1 mole PCl5 taken 0.4 moles decomposes.

For every 5 moles PCl5 taken 'X' moles decomposes.

\large \boxed{x=2}

Number of moles of PCl5 decomposed = 2

For every 2 moles of PCl5 decomposing 2 moles each of PCl3 and Cl2 are formed.

Stoichiometry
\large \mathop {PC{l_{5(g)}}}\limits^{1{\mkern 1mu} mole}
\rightleftharpoons
\large \mathop {PC{l_{3(g)}}}\limits^{1{\mkern 1mu} mole} +
\large \mathop {C{l_{2(g)}}}\limits^{1{\mkern 1mu} mole}
Initial moles5 --
Moles at equilibrium(5 - 2) 22
Equilibrium concentration
\large \frac{{3}}{0.5}
 
\large \frac{{2}}{0.5}
\large \frac{{2}}{0.5}
\large {K_C} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}}
\large {K_C} = \frac{{4 \times 4}}{6} = 2.6

 

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