A mixture of SO- and O2 under atmospheric pressure in the ratio of 2 - 1 is passed over a catalyst at 1170-C- After equilibrium has reached- the gas coming out has been found to contain 87- SO3 by volume- Calculate Kp for SO2 - 12 O2- SO3-

Let initially 2 moles of SO_2 and 1 mole of O_2 are present. #160; #160; #160; #160; #160; #160; #160; #160; #160; #160 ...

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Byju's Answer

Standard IX

Chemistry

Ways of Representing the Concentration of a Solution

A mixture of ...

Question

A mixture of $S O$ , and $O_{2}$ under atmospheric pressure in the ratio of $2 : 1$ is passed over a catalyst at $1170^{\circ} C$ . After equilibrium has reached, the gas coming out has been found to contain $87 % S O_{3}$ by volume. Calculate $K_{p}$ for $S O_{2} + \frac{1}{2} O_{2} ⇌ S O_{3}$ .

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Solution

Let initially 2 moles of $S O_{2}$ and 1 mole of $O_{2}$ are present.
$S O_{2}$
$O_{2}$ $S O_{3}$
Initial number of moles
2
1
0
Equilibrium number of moles
$2 - x$ $1 - 0.5 x$ $x$

Total number of moles $= 2 - x + 1 - 0.5 x + x = 3 - 0.5 x$
87% of these corresponds to $S O_{3}$
$(3 - 0.5 x) \times 0.87 = x$
$x = 1.82$
$S O_{2}$
$O_{2}$ $S O_{3}$
Equilibrium number of moles
0.18
0.09
1.82

$K_{p} = \frac{P_{S O_{3}}}{P_{S O_{2}} P_{O_{2}}^{0} .5} = \frac{1.82}{0.18 (0.09)^{0.5}} = 33.70 = X$
$100 X = 100 \times 33.70 = 3370$

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Similar questions

Q. A mixture of

{S O}_{2}

and

O_{2}

1

atmosphere in the ratio of

2 : 1

is passed through a catalyst at

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for attainment of equilibrium. The exit gas is found to contain

87

{S O}_{3}

by volume. Calculate

K_{p}

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S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ S O_{3} (g)

Q. The equilibrium constant

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at equilibrium.

The equilibrium constant Kp of the reaction, 2 $S O_{2}$ (g) + $O_{2}$ (g) $\leftrightharpoons$ 2 $S O_{3}$ (g) is 900 atm. at 800 K. A mixture containing $S O_{3}$ and $O_{2}$ having initial pressure of 1 and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of $S O_{2}$ gas at 800 K.