k = 0.040 for the system below at 450°c. if a reaction is initiated with 2.375 mole of cl₂ and 3.094 mole of…

Explanation:

Step1: Calculate initial concentrations

The formula for molarity $M=\frac{n}{V}$. For $Cl_2$, $n = 2.375$ mol and $V=2$ L, so $[Cl_2]_0=\frac{2.375}{2}=1.1875$ M. For $PCl_3$, $n = 3.094$ mol and $V = 2$ L, so $[PCl_3]_0=\frac{3.094}{2}=1.547$ M, and initially $[PCl_5]_0 = 0$ M.

Step2: Set up the ICE - table

Let $x$ be the change in concentration of $PCl_5$ at equilibrium.

	$PCl_5$	$PCl_3$	$Cl_2$
Initial (M)	0	1.547	1.1875
Change (M)	$x$	$-x$	$-x$
Equilibrium (M)	$x$	$1.547 - x$	$1.1875 - x$

Step3: Write the equilibrium - constant expression

The equilibrium - constant expression for the reaction $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$ is $K=\frac{[PCl_3][Cl_2]}{[PCl_5]}$. Given $K = 0.040$, we substitute the equilibrium concentrations into the expression: $0.040=\frac{(1.547 - x)(1.1875 - x)}{x}$.

Step4: Expand and simplify the equation

Expand the numerator: $(1.547 - x)(1.1875 - x)=1.547\times1.1875-1.547x - 1.1875x+x^{2}=1.8370 - 2.7345x+x^{2}$. So the equation becomes $0.040x=1.8370 - 2.7345x+x^{2}$, or $x^{2}-2.7745x + 1.8370=0$.

Step5: Solve the quadratic equation

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-2.7745$, and $c = 1.8370$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-2.7745)^{2}-4\times1\times1.8370=7.698 - 7.348=0.35$. Then $x=\frac{2.7745\pm\sqrt{0.35}}{2}=\frac{2.7745\pm0.592}{2}$. We get two solutions for $x$: $x_1=\frac{2.7745 + 0.592}{2}=1.68325$ and $x_2=\frac{2.7745 - 0.592}{2}=1.09125$. But $x_1$ is not valid because if $x = 1.68325$, then $1.547-x<0$ and $1.1875 - x<0$. So we take $x = 1.09125$.

Step6: Calculate the equilibrium concentration of $Cl_2$

$[Cl_2]=1.1875 - x$. Substitute $x = 1.09125$ into the equation, we get $[Cl_2]=1.1875-1.09125 = 0.09625$ M.

Answer:

$0.09625$ M