A sample of pure PCl 5 was introduced into an evacuated vessel at 473K. After equilibrium was attained, the concentration of PCl 5 was found to be 0·5 × 10 –1 mol L –1 . If the value of K c is 8·3 × 10 –3 , what are the concentrations of PCl 3 and Cl 2 at equilibrium? - Wired Faculty

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A sample of pure PCl₅ was introduced into an evacuated vessel at 473K. After equilibrium was attained, the concentration of PCl₅ was found to be 0·5 × 10^–1 mol L^–1. If the value of K_c is 8·3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
$PCl subscript 5 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space space space PCl subscript 3 left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis$

Solution

Let at equilibrium,

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PCl subscript 5 left parenthesis straight g right parenthesis space space space space space space space space space space space rightwards harpoon over leftwards harpoon space space space space space space space space space space space space space space space space space space space PCl subscript 3 left parenthesis straight g right parenthesis space space space plus space space space space Cl subscript 2 left parenthesis straight g right parenthesis

At eq. 0.5 x 10^-1 mol L^-1x mol L-1 x mol L-1
Applying law of chemical equilibrium

straight K subscript straight c space equals space fraction numerator open square brackets PCl subscript 3 left parenthesis straight g right parenthesis close square brackets space open square brackets Cl subscript 2 left parenthesis straight g right parenthesis close square brackets over denominator open square brackets PCl subscript 5 left parenthesis straight g right parenthesis close square brackets end fraction space space space space... left parenthesis 1 right parenthesis

Putting the values in expression (1), we have

straight K subscript straight c space equals space fraction numerator straight x squared over denominator 0.5 space cross times space 10 to the power of negative 1 end exponent end fraction space equals space 8.3 space cross times space 10 to the power of negative 3 end exponent space space left square bracket Given right square bracket straight x squared space equals space left parenthesis 8.3 space straight x space 10 to the power of negative 3 end exponent right parenthesis thin space left parenthesis 0.5 space cross times space 10 to the power of negative 1 end exponent right parenthesis space space space space equals space 4.15 space cross times space 10 to the power of negative 4 end exponent straight x space equals space square root of 4.15 space cross times space 10 to the power of negative 4 end exponent end root space equals space 2.04 space cross times space 10 to the power of negative 2 end exponent straight M space space space equals space 0.02 space straight M therefore space space open square brackets PCl subscript 3 close square brackets space equals space open square brackets Cl subscript 2 close square brackets space equals space 0.02 space straight M

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Equilibrium

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