The reaction A(g) rightleftharpoons B(g) + C( | vedclass

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(B) Initial moles: $A = a, B = 0, C = 0$.Moles at equilibrium: $A = a-x, B = x, C = x$.Total moles at equilibrium = $(a-x) + x + x = a+x$.Mole fractio

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$\frac{x^{2}}{a^{2}-x^{2}} 	imes p$

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(B) Initial moles: $A = a, B = 0, C = 0$.Moles at equilibrium: $A = a-x, B = x, C = x$.Total moles at equilibrium = $(a-x) + x + x = a+x$.Mole fractions at equilibrium: $X_{A} = \frac{a-x}{a+x}, X_{B} = \frac{x}{a+x}, X_{C} = \frac{x}{a+x}$.Partial pressures: $P_{A} = \frac{a-x}{a+x}p, P_{B} = \frac{x}{a+x}p, P_{C} = \frac{x}{a+x}p$.The equilibrium constant $K_{p}$ is given by:$K_{p} = \frac{P_{B}P_{C}}{P_{A}} = \frac{[\frac{x}{a+x}p] \cdot [\frac{x}{a+x}p]}{[\frac{a-x}{a+x}p]} = \frac{x^{2}p^{2}}{(a+x)^{2}} \cdot \frac{(a+x)}{(a-x)p} = \frac{x^{2}p}{(a+x)(a-x)} = \frac{x^{2}p}{a^{2}-x^{2}}$.

The reaction $A(g) \rightleftharpoons B(g) + C(g)$ was initiated with the amount $a$ of $A(g)$. At equilibrium, it is found that the amount of $A(g)$ remaining is $(a-x)$ at a total pressure of $p$. The equilibrium constant $K_{p}$ of the reaction can be calculated from the expression:

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