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

Raoult's Law

For the react...

Question

For the reaction, $H_{2} (g) + C O_{2} (g) ⇋ C O (g) + H_{2} O (g)$ , if the initial concentration of $[H_{2}] = [C O_{2}] = 1$ and $x$ mol/litre of $H_{2}$ is consumed at equilibrium, the correct expression of $K_{p}$ is:

\frac{x^{2}}{{(1 - x)}^{2}}

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

\frac{{(1 + x)}^{2}}{{(1 - x)}^{2}}

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

\frac{x^{2}}{{(2 + x)}^{2}}

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

\frac{x^{2}}{1 - x^{2}}

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

Open in App

Solution

The correct option is A $\frac{x^{2}}{{(1 - x)}^{2}}$
$E x p l a n a t i o n :$
$H_{2 (g)} + C O_{2} ⇌ C O + H_{2} O_{(g)}$
Initial concentration: $1$ $1$ $0$ $0$
At equilibrium: $(1 - x)$ $(1 - x)$ $x$ $x$

$K_{P} = \frac{P_{C O} \times P_{H_{2} O}}{P_{H_{2}} P_{C O_{2}}} = \frac{x \times x}{(1 - x) \times (1 - x)} = \frac{x^{2}}{(1 - x)^{2}}$

Hence the correct answer is option $A$ .

Suggest Corrections

Similar questions

H_{2} (g) + C O_{2} (g) ⇌ C O_{2} (g) + H_{2} O (g)

[H_{2}] = [C O_{2}]

x m o l / L

K_{p}

X_{2} + Y_{2} ⇌ 2 X Y

X_{2}

Y_{2}

X_{2}

Y_{2}

m o l^{- 1}

f (x) = x^{2} + \frac{x^{2}}{(1 + x^{2})} + \frac{x^{2}}{(1 + x^{2})^{2}} + \dots + \frac{x^{2}}{(1 + x^{2})^{n}} + \dots

x = 0

f (x) = {tan}^{- 1} (\frac{1 - \sqrt{x^{2} - 1}}{\sqrt{x^{2} + 2 \sqrt{x^{2} - 1}}}) + {sin}^{- 1} (\frac{\sqrt{x^{2} - 1}}{| x |})

| x | \geq 1,

tan (f (x)) = | x^{2} - 2 |

X_{2} + Y_{2} ⇌ 2 X Y

1

X_{2}

2

Y_{2}

2

X_{2}

Y_{2}

[X Y] = 0.6 {m o l}^{- 1}

Join BYJU'S Learning Program