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Standard XII

Chemistry

Reaction Quotient

For the react...

Question

For the reaction :-
$2 C a S O_{4} (s) ⇌ 2 C a O (s) + 2 S O_{2} (g) + O_{2} (g)$ ,
$K_{p} = 1.45 \times 10^{- 5} {(a t m)}^{3}$ at $1625 K$ . If mole fraction of $S O_{2}$ is 0.4 at equilibrium, the dissociation pressure of $C a S O_{4}$ in $m m$ of $H g$ at $1625 K$ is $X$ atm. $100 X$ is___________.

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Solution

$K_{p} = P_{S O_{2}}^{2} \times P_{O_{2}}$
$1.45 \times 10^{- 5} = (0.4 P)^{2} \times 0.6 P$
$P^{3} = 1.51 \times 10^{- 4}$
$P = 0.053 a t m$
$P = 760 \times 0.053 = 40.4 m m$
$X = 40.47$
$100 X = 4047$

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For the reaction :- 2CaSO4(s)⇌2CaO(s)+2SO2(g)+O2(g),Kp=1.45×10−5(atm)3 at 1625K. If mole fraction of SO2 is 0.4 at equilibrium, the dissociation pressure of CaSO4 in mm of Hg at 1625K is X atm. 100X is___________.

Kp=P2SO2×PO21.45×10−5=(0.4P)2×0.6PP3=1.51×10−4P=0.053atmP=760×0.053=40.4mmX=40.47100X=4047

$K_{p} = P_{S O_{2}}^{2} \times P_{O_{2}}$
$1.45 \times 10^{- 5} = (0.4 P)^{2} \times 0.6 P$
$P^{3} = 1.51 \times 10^{- 4}$
$P = 0.053 a t m$
$P = 760 \times 0.053 = 40.4 m m$
$X = 40.47$
$100 X = 4047$