$0.5$ mole of hydrogen and $0.5$ mole of iodine react in a $10$ litre evacuated vessel at $448^{o} C$ ; hydrogen iodine formed. The equilibrium constant, $K_{c}$ for the reaction is $50$ .
(a) Calculate the number of moles of iodine which remain unreacted at equilibrium.
(b) What is the value of $K_{p}$ ?

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Solution

(a) $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$
Initial moles $0.5$ $0.5$ $0.0$
No. of moles
at equilibrium $(0.5 - x)$ $(0.5 - x)$ $2 x$
$K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]}$
$K_{c} = \frac{{(2 x m o l / 10 L)}^{2}}{((0.5 - x) m o l / 10 L) \times ((0.5 - x) m o l / 10 L)} = 50$
$\frac{2 x}{(0.5 - x)} = \sqrt{50} = 7.07$
$\frac{x}{(0.5 - x)} = 3.54$
$\frac{x}{3.54} = (0.5 - x)$
$0.2828 x = (0.5 - x)$
$1.2828 x = 0.5$
$x = 0.39$ moles
Number of moles of $I_{2}$ at equilibrium $= (0.5 - x) = (0.5 - 0.39) = 0.11$ mole
(b)
$K_{p} = K_{c} {(R T)}^{Δ n}$
As $Δ n = 2 - (1 + 1) = 0$
$K_{p} = K_{c} {(R T)}^{0} = K_{c} = 50$

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0.5 mole of hydrogen and 0.5 mole of iodine react in a 10 litre evacuated vessel at 448oC; hydrogen iodine formed. The equilibrium constant, Kc for the reaction is 50.(a) Calculate the number of moles of iodine which remain unreacted at equilibrium.(b) What is the value of Kp?