Question

The reaction, $2A\left(g\right)+B\left(g\right)\rightleftharpoons 3C\left(g\right)+D\left(g\right)$ is begun with the concentrations of $A$ and $B$ both at an initial value of $1.00M$. 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. $\left[\right(0.75{)}^3\left(0.25\right)]÷[\left(1.00{)}^2\right(1.00\left)\right]$
• B. $\left[\right(0.75{)}^3\left(0.25\right)]÷[\left(0.50{)}^2\right(0.75\left)\right]$
• C. $\left[\right(0.75{)}^3\left(0.25\right)]÷[\left(0.50{)}^2\right(0.25\left)\right]$
• D. $\left[\right(0.75{)}^3\left(0.25\right)]÷[\left(0.75{)}^2\right(0.25\left)\right]$

Answer 1

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

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

Initially	1	1	0	0
At Eqm.	1 - 2x = 1- 0.5	1 - x =1 -0.25	3x = 0.75	x = 0.25

$K=\frac{\left[C{]}^3\right[D]}{\left[A{]}^2\right[B]}$

$K=\frac{\left(0.75{)}^3\right(0.25)}{\left(0.50{)}^2\right(0.75)}$