Question

The reaction, $2A\left(g\right)+B\left(g\right)\rightleftharpoons 3C\left(g\right)+D\left(g\right)$, is begun with concentration of $A$ and $B$ both at initial value of $1M$. When equilibrium is reached, the concentration of $D$ is measured and found to be $0.25M$. The value for the equilibrium constant for this reaction is given by the expression:

• A. $\left[{\left(0.75\right)}^3\left(0.25\right)\right]÷\left[{\left(1.00\right)}^2\left(1.00\right)\right]$
• B. $\left[{\left(0.75\right)}^3\left(0.25\right)\right]÷\left[{\left(0.50\right)}^2\left(0.75\right)\right]$
• C. $\left[{\left(0.75\right)}^3\left(0.25\right)\right]÷\left[{\left(0.50\right)}^2\left(0.25\right)\right]$
• D. $\left[{\left(0.75\right)}^3\left(0.25\right)\right]÷\left[{\left(0.75\right)}^2\left(0.25\right)\right]$

Answer 1

The correct option is B $\left[{\left(0.75\right)}^3\left(0.25\right)\right]÷\left[{\left(0.50\right)}^2\left(0.75\right)\right]$

$2A\left(g\right)+B\left(g\right)\rightleftharpoons 3C\left(g\right)+D\left(g\right)$

$At{t}_0$ $1$ $1$ $0$ $0$
$Att$ $1-2x$ $1-x$ $3x$ $x$

Given,

$\left[D\right]=0.25M$

$K=\frac{\left[C{]}^3\right[D]}{\left[A{]}^2\right[B]}=\frac{{\left(0.75\right)}^3\left(0.25\right)}{{\left(0.5\right)}^2\left(0.75\right)}$

Hence, option B is correct.