CameraIcon

SearchIcon

MyQuestionIcon

128

You visited us 128 times! Enjoying our articles? Unlock Full Access!

Reaction Quotient

At 700 K, ...

Question

At $700 K$ , ${C O}_{2}$ and $H_{2}$ react to form $C O$ and $H_{2} O$ . For this process $K_{c}$ is $0.11$ . If a mixture of $0.45$ mole of ${C O}_{2}$ and $0.45$ mole of $H_{2}$ is heated at $700 K$ .
(i) Find out the amount of each gas at equilibrium state.
(ii) After equilibrium is reached another $0.34 m o l e$ of ${C O}_{2}$ and $0.34 m o l e$ of $H_{2}$ are added to the reaction mixture. Find the composition of the mixture at the new equilibrium state.

Open in App

Solution

$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$
Initial no. of moles: $0.45$ $0.45$ $0$ $0$
No. of moles at $e q^{m}$ : ( $0.45 - x$ ) ( $0.45 - x$ ) $x$ $x$
Applying law of mass action,
$K_{c} = \frac{[C O] [H_{2} O]}{[C O_{2}] [H_{2}]} = \frac{x \times x}{(0.45 - x) (0.45 - x)} = 0.11$
So, $\frac{x}{(0.45 - x)} = 0.33$
$x = 0.11$
At equilibrium, $[C O_{2}] = [H_{2}] = 0.34 m o l e$
$[C O] = [H_{2} O] = 0.11 m o l e$
(ii) $C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$
Initial moles $0.34$ $0.34$ $0.11$ $0.11$
$+ 0.34$ $+ 0.34$ $-$ $-$
$= 0.68$ $= 0.68$ $0.11$ $0.11$
At equilibrium:( $0.68 - y$ ), ( $0.68 - y$ ), ( $0.11 + y$ ), ( $0.11 + y$ )
$K_{c} = \frac{[C O] [H_{2} O]}{[C O_{2}] [H_{2}]} = \frac{(0.11 + y) (0.11 + y)}{(0.68 - y) (0.68 - y)} = 0.11$
or $y = 0.086$
At equilibrium, $[C O_{2}] = [H_{2}] = 0.11 + 0.086 = 0.196 m o l e$
$[C O] = [H_{2} O] = 0.68 - 0.086 = 0.594 m o l e$

Suggest Corrections

thumbs-up

0

Similar questions

700 K

hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is

5 \times 10^{8}

. Calculate the amount of

H_{2}, {B r}_{2}

H B r

at equilibrium if a mixture of

0.6 m o l

H_{2}

0.2 m o l

of bromine is heated to

700 K

Q. A mixture of one mole of

C O_{2}

and one mole of

H_{2}

attains equilibrium at

250^{\circ} C

C O_{2} + H_{2} \to C O + H_{2} O

K_{p}

, if the analysis of final reaction mixture shows 0.16 volume percent of CO :

675 K

H_{2} (g)

C O_{2} (g)

C O (g)

H_{2} O (g), K_{p}

for the reaction is

0.16

. If a mixture of

0.25

H_{2}

0.25

C O_{2}

675 K

, mole percentage of

C O (g)

in equilibrium mixture is:

Q. At 100 K, then value of

K_{c}

for the reaction

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

3.0 \times 10^{2}

. Calculate equilibrium concentrations of

H_{2} O, C O_{2}

H_{2}

in the reaction mixture obtained by heating 6.0 mole of steam and an excess of solid carbon in a 5.0 L container. What is the molar composition of the equilibrium mixture?

Q. An equilibrium mixture of

N_{2}

H_{2}

N H_{3}

700 K

0.036 M N_{2}

0.15 M

H_{2}

. At this temperature,

K_{c}

for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

0.29

. What is the concentration of

N H_{3}

?

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equilibrium Constants

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Reaction Quotient

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon