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Relation between Kp and Kc

At 800∘ C, th...

Question

At $800^{\circ}$ C, the following equilibrium is established as
$F_{2} (g) ⇌ 2 F (g)$
The composition of equilibrium may be determined by measuring the rate of effusion of the mixture through a pin hole. It is found that at $800^{\circ}$ C and 1 atm mixture effuses 1.6 times as fast as $S O_{2}$ effuses under the similar conditions. (Atomic weight of F =19 amu). What is the value of $K_{p}$ ?

A

0.315 atm

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B

0.685 atm

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C

0.460 atm

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D

1.490 atm

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Solution

The correct option is D 1.490 atm
$\frac{r_{m i x}}{r_{S O_{2}}}$ = $(\frac{M_{S O_{2}}}{M_{m i x}})^{\frac{1}{2}}$
2.56 = $\frac{64}{M_{m i x}}$
$M_{m i x}$ = 25
Let's say mole fraction of $F_{2}$ is $x$
25 = $\frac{38 \times x + (1 - x) \times 19}{1}$
$x = 0.315$
$K_{p} = \frac{(P_{F})^{2}}{P_{F_{2}}}$ = $\frac{(0.685)^{2}}{0.315} = 1.49 a t m$

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