Question

$0.5mole$ of $H_2\left(g\right)$ and $1.0mole$ of $HI\left(g\right)$ (but no $I_2)$ are added to a $10L$ vessel and allowed to reach equilibrium according to the following reaction: $H_2+I_2\rightleftharpoons 2HI\left(g\right)$. If $x$ is the equilibrium moles of $I_2\left(g\right)$, the correct equilibrium constant expression is__________.

• A. $\frac{x(0.5-x)}{(1+2x{)}^2}$
• B. $\frac{(2-x{)}^5}{(0.5+x{)}^2}$
• C. $\frac{x(2+x{)}^2}{(0.5+x)}$
• D. $\frac{(1-2x{)}^2}{(0.5+x)x}$

Answer 1

Answer: (A)

$\frac{x(0.5-x)}{(1+2x{)}^2}$

Given no. of initial moles of $H_2\to 0.5moles;HI\to 1moles$

$H_2+I_2⟶2HI$

at $t=0;0.5;0;1$

at $e{q}^m;0.5-x;x;1+2x$

$k=\frac{x(0.5-x)}{(1+2x{)}^2}$

$k=$equilibrium constant.