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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:$K_{p} = \frac{P_{B}P_{C}}{P_{A}} = \frac{[\frac{x}{a+x}p] \cdot [\frac{x}{a+x}p]}{[\frac{a-x}{a+x}p]} = \frac{x^{2}p^{2}}{(a+x)^{2}} \cdot \frac{(a+x)}{(a-x)p} = \frac{x^{2}p}{(a+x)(a-x)} = \frac{x^{2}p}{a^{2}-x^{2}}$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

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:

Similar Questions

For the reaction,$H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$,which of the following relations is correct?

The equilibrium $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ shows that $K_P$ (in $atm$) is double the value of $K_C$ (in $mol/L$) at a particular temperature $T$. Then,$T$ is $...... \ K$.

At $300 \ K$,$K_C$ for the reaction $A_2B_{2(g)} \rightleftharpoons A_{2(g)} + B_{2(g)}$ is $100 \ mol \ L^{-1}$. What is its $K_p$ (in $atm$) at the same temperature? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

$A$ schematic plot of $\ln K_{eq}$ versus inverse of temperature $(1/T)$ for a reaction is shown below. The reaction must be

Vedclass Test Series

Exam Paper Generator

Online Exam Module