Question

1. equilibrium is established in a 2.50 l flask at 250 °c for the following reaction: ( pcl_{5}(g)

ightleftharpoons pcl_{3}(g)+cl_{2}(g)) ( k_{c}=3.8\times10^{-2}). what are the equilibrium concentrations of all three species if 0.550 mol of ( pcl_{5}) and 0.550 mol of ( pcl_{3}) were introduced into the empty flask?

Explanation:

Step1: Calculate initial concentrations

The initial concentration of $PCl_5$ is $c_{PCl_5}=\frac{0.550\ mol}{2.50\ L}= 0.220\ M$, and the initial concentration of $PCl_3$ is $c_{PCl_3}=\frac{0.550\ mol}{2.50\ L}=0.220\ M$. The initial concentration of $Cl_2$ is $0\ M$. Let the change in concentration of $PCl_5$ be $x\ M$. Then the change in concentration of $PCl_3$ is $x\ M$ and the change in concentration of $Cl_2$ is $x\ M$.

Step2: Set up the equilibrium - constant expression

The equilibrium - constant expression for the reaction $PCl_5(g)
ightleftharpoons PCl_3(g)+Cl_2(g)$ is $K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$. At equilibrium, $[PCl_5]=0.220 - x$, $[PCl_3]=0.220 + x$, and $[Cl_2]=x$. Substituting these into the equilibrium - constant expression gives $3.8\times10^{-2}=\frac{(0.220 + x)x}{0.220 - x}$.

Step3: Rearrange the equation

Cross - multiply to get $3.8\times10^{-2}(0.220 - x)=(0.220 + x)x$. Expand both sides: $3.8\times10^{-2}\times0.220-3.8\times10^{-2}x = 0.220x+x^{2}$. So, $0.00836-0.038x=0.220x + x^{2}$. Rearrange to the standard quadratic form $x^{2}+0.258x - 0.00836 = 0$.

Step4: Solve the quadratic equation

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b = 0.258$, and $c=-0.00836$. First, calculate the discriminant $\Delta=b^{2}-4ac=(0.258)^{2}-4\times1\times(-0.00836)=0.066564 + 0.03344=0.100004$. Then $x=\frac{-0.258\pm\sqrt{0.100004}}{2}$. We take the positive root since concentration cannot be negative. $x=\frac{-0.258 + 0.3162}{2}=\frac{0.0582}{2}=0.0291\ M$.

Step5: Calculate equilibrium concentrations

$[PCl_5]=0.220 - x=0.220-0.0291 = 0.191\ M$.
$[PCl_3]=0.220 + x=0.220 + 0.0291=0.249\ M$.
$[Cl_2]=x = 0.0291\ M$.

Answer:

$[PCl_5]=0.191\ M$, $[PCl_3]=0.249\ M$, $[Cl_2]=0.0291\ M$