Question

Calculate the

K_{p}

and

K_{c}

of the reaction:

I_{2} ⇌ 2 I

when

1.513 \times 10^{- 3}

mole

of iodine is heated to

1073 K

, if its vapour occupies a volume of

249.3 \times 10^{- 6} m^{3}

at

5.81 \times 10^{4} Pa

.

Accepted Answer

Given,

$I_{2} (g) ⇌ 2 I (g) t = 0 1 0 t = eq 1 - α 2 α$

The moles of iodine molecules which are heated to convert into iodine atom is $α = 1.513 \times 10^{- 3} mole$

Total moles at equilibrium is $= 1 - α + 2 α = 1 + α$

$K_{p} = \frac{{P_{I}}^{2}}{P_{I_{2}}}$

To calculate the $K_{P}$ , we need the partial pressure of each at equilibrium.

Partial pressure $=$ $Mole fraction \times Total pressure$

The mole fraction of $I_{2}$ at equilibrium $= \frac{moles of I_{2}}{Total moles} = \frac{1 - α}{1 + α} = \frac{1 - 1.53 \times 10^{- 3}}{1 + 1.53 \times 10^{- 3}} = 0.9969$

The mole fraction of $I$ at equilibrium $= \frac{moles of I}{Total moles} = \frac{2 α}{1 + α} = \frac{2 \times 1.53 \times 10^{- 3}}{1 + 1.53 \times 10^{- 3}} =$ $0.0031$

Total vapour pressure at equilibrium is given $5.81 \times 10^{4} pa$ .

$K_{p} = \frac{{P_{I}}^{2}}{P_{I_{2}}} = \frac{(x_{I} \times P_{T})^{2}}{x_{I_{2}} \times P_{T}} = \frac{(0.0031 \times 5.81 \times 10^{4})^{2}}{0.9969 \times 5.81 \times 10^{4}} = 0.01113$

Now, calculate the $K_{c}$ .

We know the relation between the $K_{p}$ and $K_{c}$ is given by

$K_{p} = K_{c} (RT)^{∆ n_{g}} or K_{c} = \frac{K_{p}}{(RT)^{∆ n_{g}}}$

Here, $∆ n_{g} =$ No. of gaseous moles in product $-$ No. of gaseous moles is reactant

$∆ n_{g} = 2 - 1 = 1$

Putting the values in the above expression,

$K_{c} = \frac{0.0113}{(0.0821 \times 1073)} = 0.1264$

So, the value of $K_{p} = 0.0113$ and $K_{p} = 0.1264$ .

Calculate the $K_{p}$ and $K_{c}$ of the reaction: $I_{2} ⇌ 2 I$ when $1.513 \times 10^{- 3}$ $mole$ of iodine is heated to $1073 K$ , if its vapour occupies a volume of $249.3 \times 10^{- 6} m^{3}$ at $5.81 \times 10^{4} Pa$ .

Important Questions on Miscellaneous Problems for Revision

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Calculate the Kp and Kc of the reaction: I2⇌2I when 1.513×10-3 mole of iodine is heated to 1073 K, if its vapour occupies a volume of 249.3×10-6 m3 at 5.81×104 Pa.

Important Questions on Miscellaneous Problems for Revision

Learn and Prepare for any exam you want !

Calculate the $K_{p}$ and $K_{c}$ of the reaction: $I_{2} ⇌ 2 I$ when $1.513 \times 10^{- 3}$ $mole$ of iodine is heated to $1073 K$ , if its vapour occupies a volume of $249.3 \times 10^{- 6} m^{3}$ at $5.81 \times 10^{4} Pa$ .