A mixture of one mole of CO2 and one mole of H2 attains equilibrium at 250-C and 0-1 atm- CO2-H2- CO-H2O Calculate Kp- if the analysis of final reaction mixture shows 0-16 volume percent of CO -

The correct option is C 1.03× 10^ 5 The equilibrium reaction is CO_2+H_2→ CO+H_2O .The following table lists various quantities. CO_2 , #160; ...

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Byju's Answer

Standard XII

Chemistry

Relation between Kp and Kc

A mixture of ...

Question

A mixture of one mole of $C O_{2}$ and one mole of $H_{2}$ attains equilibrium at $250^{\circ} C$ and 0.1 atm.
$C O_{2} + H_{2} \to C O + H_{2} O$
Calculate $K_{p}$ , if the analysis of final reaction mixture shows 0.16 volume percent of CO :

2.06 \times 10^{5}

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10.3 \times 10^{5}

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1.03 \times 10^{- 5}

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2.06 \times 10^{- 5}

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Solution

The correct option is C $1.03 \times 10^{- 5}$
The equilibrium reaction is $C O_{2} + H_{2} \to C O + H_{2} O$ .
The following table lists various quantities.

$C O_{2}$ , $H_{2}$ ,
$C O$
, $H_{2} O$
Initial moles:
1
1
0
0
Moles at equilibrium:
$1 - x$
$1 - x$
$x$
$x$
Total number of moles $= 1 - x + 1 - x + x + x = 2$ .
For $C O$ , mole % is equal to volume %.
Thus $\frac{x}{2} \times 100 = 0.16$ or $x = 0.0032$ .
As, $Δ n = 0$ , $K_{c} = K_{p}$
$= \frac{x^{2}}{(1 - x)^{2}}$
$= \frac{(0.0032)^{2}}{(1 - 0.0032)^{2}}$
$= 1.03 \times 10^{- 5}$ .

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	$C O_{2}$	, $H_{2}$ ,	$C O$	, $H_{2} O$
Initial moles:	1	1	0	0
Moles at equilibrium:	$1 - x$	$1 - x$	$x$	$x$

A mixture of one mole of CO2 and one mole of H2 attains equilibrium at 250∘C and 0.1 atm.CO2+H2→CO+H2O Calculate Kp, if the analysis of final reaction mixture shows 0.16 volume percent of CO :

A mixture of one mole of $C O_{2}$ and one mole of $H_{2}$ attains equilibrium at $250^{\circ} C$ and 0.1 atm.
$C O_{2} + H_{2} \to C O + H_{2} O$
Calculate $K_{p}$ , if the analysis of final reaction mixture shows 0.16 volume percent of CO :