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

Nernst Equation

Kc for the re...

Question

$K_{c}$ for the reaction $S O_{2 (g)} + \frac{1}{2} O_{2 (g)} ⇌ S O_{3 (g)}$ is $61.7$ at $60^{o}$ C. What is its unit? Calculate $K_{p}$ for the reaction and write its unit?

Open in App

Solution

Unit of $K_{c}$ for given reaction is $l m o l^{- 1}$ and with given coefficients the unit is $(l m o l^{- 1})^{1 / 2}$ .
$K_{p} = K_{c} (R T)^{Δ n}$
For given reaction $K_{c} = 61.7, R = 0.0821, T = 60 + 273 = 333$ K and $Δ_{n} =$ number of moles of gaseous products $-$ Number of moles of gaseous reactants

$= [1 - (1 + \frac{1}{2})] = - \frac{1}{2}$

$∴ K_{p} = 61.7 (0.0821 \times 333)^{- \frac{1}{2}}$

$= \frac{61.7}{\sqrt{27.34}} = 11.8$ atm.

Suggest Corrections

Similar questions

(K_{p} / K_{c})

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

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

K_{c}

1.7 \times 10^{12}

300 K

300 K

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

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

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

K_{p}

32 a t m^{- 1 / 2}

800 K

Δ H

- 187.9 k J

{m o l}^{- 1}

900 K

Δ H

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

K_{p} = K_{C} (R T)^{x}

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

K_{P} = K_{C} {(R T)}^{x}

x

Join BYJU'S Learning Program