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Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

In the equili...

Question

In the equilibrium reaction, $2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)$ , the partial pressures of $S O_{2}, O_{2}$ and $S O_{3}$ are $0.662$ atm, $0.101$ atm and $0.662$ atm respectively. Find the value of $K_{p}$ (in atm $^{- 1})$ :

0.303

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9.9

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0.606

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0.110

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Solution

The correct option is A $9.9$
The equilibrium reaction is given below:

$2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)$
The expression for the equilibrium constant, $K_{p} = \frac{P_{S O_{3}}^{2}}{P_{S O_{2}}^{2} \times P_{O_{2}}}$
Substituting values in the above expression, we get
$K_{p} = \frac{{0.662}^{2}}{{0.662}^{2} \times 0.101} = 9.9$ atm $^{- 1}$

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Similar questions

Q. At T(K), the partial pressures of

S O_{2}, O_{2}

and

S O_{3}

are 0.662, 0.100 and 0.331 atm, respectively for the reaction

{2 S O_{2}}_{(g)} + {O_{2}}_{(g)} ⇌ {2 S O_{3}}_{(g)}

at equilibrium. What is the partial pressure in atm, of

O_{2}

when the equilibrium partial pressures of

S O_{2}

and

S O_{3}

are equal at the same temperature ?

Q. In the equilibrium,

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g),

the equilibrium partial pressure of

S O_{2}, O_{2}

and

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0.6, 0.1

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Q. In the equilibrium

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

, the partial pressure of

S O_{2}, O_{2}

and

S O_{3}

are

0.662, 0.10

and

0.331

atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentrations of

S O_{2}

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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.

Q. The equilibrium constant

K_{p}

of the reaction:

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

is 900 atm at 800 K. A mixture containing

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and

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having initial partial pressure of 1 and 2 atm. respectively is heated at constant volume to equilibriate. Calculate the partial pressure of

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In the equilibrium reaction, 2SO2(g)+O2(g)⇋2SO3(g), the partial pressures of SO2,O2 and SO3 are 0.662 atm, 0.101 atm and 0.662 atm respectively. Find the value of Kp (in atm−1):

The correct option is A 9.9The equilibrium reaction is given below:2SO2(g)+O2(g)⇋2SO3(g)The expression for the equilibrium constant, Kp=P2SO3P2SO2×PO2Substituting values in the above expression, we getKp=0.66220.6622×0.101=9.9 atm−1

In the equilibrium reaction, $2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g)$ , the partial pressures of $S O_{2}, O_{2}$ and $S O_{3}$ are $0.662$ atm, $0.101$ atm and $0.662$ atm respectively. Find the value of $K_{p}$ (in atm $^{- 1})$ :