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Chemical Equilibrium question

2004 · Shift 0 · Q4

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Chemical Equilibrium
2004 · Shift 0 · Q4

Chemical Equilibrium question

2004 · Shift 0 · Q4

JEE MainChemistryChemical EquilibriumMCQ+4 / −1

What is the equilibrium expression for the reaction

P_4

(s) + 5

O_2

\leftrightharpoons

P_4O_{10}

(s)?

A
Kc = [ $P_4O_{10}$ ] / 5[ $P_4$ ] [ $O_2$ ]
B
Kc = 1/[ $O_2$ ]5
C
Kc = [ $P_4O_{10}$ ] / [ $P_4$ ] [ $O_2$ ]5
D
Kc = [ $O_2$ ]5

View written solutionFree

Correct answer: B

Write the reaction clearly

$P_4(s) + 5O_2(g) \rightleftharpoons P_4O_{10}(s)$

Rule for equilibrium expressions

For a heterogeneous equilibrium, the activities of pure solids are taken as $1$ .

So:

$P_4(s)$ is a solid $\Rightarrow$ omitted from $K_c$
$P_4O_{10}(s)$ is a solid $\Rightarrow$ omitted from $K_c$
Only $O_2(g)$ appears in the equilibrium expression

Form the equilibrium expression

Using products over reactants:

$K_c = \frac{1}{[O_2]^5}$

because the solid product contributes $1$ and the solid reactant also contributes $1$ .

Match with the options

A: Incorrect, because solids should not appear in $K_c$
B: Correct, since it represents $K_c = \frac{1}{[O_2]^5}$
C: Incorrect, because solids are included wrongly
D: Incorrect, exponent is in numerator instead of denominator

Final answer

$\boxed{K_c = \frac{1}{[O_2]^5}}$

So the correct option is B.

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