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Question

Kc for PCl5(g)PCl3(g)+Cl2(g) is 0.45 at 250C. How many mole of PCl5 must be added to a 3 - litre flask to obtain a Cl2 concentration of 0.15 M?

A
0.45
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B
0.6
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C
1.35
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D
1.8
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Solution

The correct option is B 0.6
Here Kc=0.45.[Cl2]=0.15M
No. of moles of Cl2 at equrilibrium concentration is 0.15×3=0.45moles.
Let the no of moles of PCl5 added is 'x'
PCl5PCl3+Cl2

Then,
(x0.453) will be the equilibrium concentration of PCl5
Kc=[PCl3][Cl2][PCl5]=0.45
0.15×0.15(x0.453)=0.45
x=0.6moles

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