Question

For the decomposition of phosphorous pentachloride to phosphorous trichloride and chlorine at 400K the KC is 1.1x10-2. Given that 1.0g of phosphorous pentachloride is added to a 250mL reaction flask, find the percent decomposition after the system has reached equilibrium.

Answer 1

Answer:

$\% Dec=48.0\%$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Thus, the initial concentration of phosphorous pentachloride is:

$[PCl_5]_0=\frac{1.0g*\frac{1mol}{208.24g} }{0.250L} =0.0192M$

In such a way, for the equilibrium, the change $x$ is introduced and the law of mass action is written as shown below:

$Kc=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{(x)(x)}{(0.0192-x)} =1.1x10^{-2}$

Hence, solving for $x$ we obtain:

$x_1=-0.021\\x_2=0.010$

Obviously, the solution is $x_2=0.010$, thus, the equilibrium concentration of phosphorous pentachloride is:

$[PCl_5]_{eq}=0.0192M-0.01M=0.00921M$

In such a way, the percent of decomposition is:

$\% Dec=\frac{[PCl_5]_{eq}}{[PCl_5]_0}*100\% =\frac{0.00921M}{0.0192M} *100\%\\\% Dec=48.0\%$

Best regards.

Answer 2

Answer:

% = 47.92%

Explanation:

To do this, let's write the equilibrium reaction for this:

PCl₅ <-------> PCl₃ + Cl₂   Kc = 1.1x10⁻²

The percent of decomposition is a number that indicates how much of the initial reactant was decomposed in the reaction. To know this number, we need to know how much of the initial reactant have left after the equilibrium is reached.

First, let's calculate the initial moles of PCl₅. The reported molar mass of PCl₅  is 208.24 g/mol so the moles:

n = 1/208.24 = 0.0048 moles

We have the moles, let's calculate the concentration using the 250 mL flask:

M = 0.0048 / 0.250 = 0.0192 M

Now that we have the concentration, let's do an ICE chart for this reaction:

         PCl₅ <-------> PCl₃ + Cl₂   Kc = 1.1x10⁻²

i)        0.0192            0        0

c)         -x                   +x      +x

e)     0.0192-x           x         x

The equilibrium expression (Kc) would be:

Kc = [PCl₃] [Cl₂] / [PCl₅]

Replacing the above data we have:

1.1*10⁻² = x² / (0.0192-x)      solving for x:

1.1*10⁻²(0.0192 - x) = x²

2.112*10⁻⁴ - 1.1*10⁻² = x²

x² + 1.1*10⁻²x - 2.112*10⁻⁴ = 0   -----> a = 1;  b = 1.1*10⁻²;  c = -2.112x10⁻⁴

Using the general equation for solving x in a quadratic equation we have:

x = -b ±√(b² - 4ac) / 2a

x = -1.1*10⁻² ±√(1.1x10⁻²)² + 4*1*2.112*10⁻⁴ / 2*1

x = -1.1*10⁻² ±√(9.658*10⁻⁴) / 2

x = -1.1*10⁻² ± 3.11x10⁻² / 2

x₁ = -1.1*10⁻² + 3.11x10⁻² / 2 = 0.010

x₂ = -1.1*10⁻² - 3.11x10⁻² / 2 = -0.02105

As x₁ is positive, this would be the correct value of x. This value corresponds to the concentration of PCl₃ and Cl₂:

[PCl₃] = [Cl₂] = 0.010 M

And the concentration of PCl₅ would be:

[PCl₅] = 0.0192 - 0.010 = 0.0092 M

Finally the percent decomposition is:

% = 0.0092 / 0.0192 * 100

% = 47.92 %

This would be the % decomposition for this reaction