You visited us 1 times! Enjoying our articles? Unlock Full Access!

Types of Equillibrium

The equilibri...

Question

The equilibrium mixture for $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ present in 1L vessel at $600^{p} C$ contains 0.50, 0.12, and 5.0 mole of $S O_{2}, O_{2} a n d S O_{3}$ respectively.
Also calculate $K_{p}$

11.62 a t m^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

21.62 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

31.62 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

18.62 a t m^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $11.62 a t m^{- 1}$
The equilibrium reaction is $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g) .$
The expression for th eequilibrium constant is $K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
Substitute values in the above expression.
$K_{c} = \frac{(5)^{2}}{(0.5)^{2} \times 0.12} = 833.33 m o l^{- 1} L$
$K_{p} = K_{c} (R T)^{△ n_{g}}$
where $△ n_{g}$ = (total gas molecules in product) - (total gas molecule in the reactant)
In this case $△ n_{g}$ = 2-3 = -1
$K_{p} = K_{c} (R T)^{- 1} = \frac{833.33}{0.0821 \times 873} = 11.62 a t m^{- 1}$

Suggest Corrections

Similar questions

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

600^{o} C

S O_{2}, O_{2} and S O_{3}

K_{C}

600^{o} C .

{2 S O}_{2} + O_{2} ⇌ {2 S O}_{3}

S O_{2}, O_{2}

S O_{3}

K_{C}

600^{0} C

K_{P}

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

600^{o} C

S O_{2}, O_{2} and S O_{3}

K_{C}

600^{o} C .

600^{\circ} C

S O_{2}

O_{2}

S O_{3}

K_{c}

600^{\circ} C

K_{p}

O_{2}

600^{\circ} C

S O_{3}

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

S O_{3}

S O_{2}

O_{2}

Join BYJU'S Learning Program