Question

For the reaction,
$2A\left(g\right)+B\left(g\right)\rightleftharpoons 3C\left(g\right)+D\left(g\right)$
two moles each of $A$ and $B$ were taken into a flask. The following must always be true when the system attained equilibrium:

• A. $\left[A\right]=\left[B\right]$
• B. $\left[A\right]<\left[B\right]$
• C. $\left[B\right]=\left[C\right]$
• D. $\left[A\right]>\left[B\right]$

Answer 1

The correct option is B $\left[A\right]<\left[B\right]$

For the reaction,
$\underset{22-2x}{2A\left(g\right)}+\underset{22-x}{B\left(g\right)}\rightleftharpoons \underset{03x}{3C\left(g\right)}+\underset{0x}{D}\left(g\right)$

2 moles each of $A$ and $B$ were taken into a flask. The consumption of A is twice of that of B.

Hence, ${\left[A\right]}_{eqm}<{\left[B\right]}_{eqm}$

Also from the reaction stoichiometry,

$-\frac{1}{2}\frac{d}{dt}\left[A\right]=-\frac{d}{dt}\left[B\right]$
or
$-\frac{d}{dt}\left[A\right]=-2\frac{d}{dt}\left[B\right]$

The rate of consumption of A is twice the rate of consumption of B.

So the equilibrium concentration of A will be less than the equilibrium concentration of B.

Option B is correct.