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3 years ago

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Consider the following reaction: COCl2(g) ⇌ CO(g) + Cl2(g) A reaction mixture initially contains 1.6 M COCl2. Determine the equi

librium concentration of CO if Kc for the reaction at this temperature is 8.33 × 10-4 (Hint: Note the size of Kc).

1 answer:

professor190 [17]3 years ago

5 0

Answer:

The equilibrium concentration of CO is 0.0361 M

Explanation:

Step 1: Data given

Kc = 8.33 *10^-4

Molarity of COCl2 = 1.6 M

Step 2: The balanced equation:

COCl2(g) ⇌ CO(g) + Cl2(g)

Step 3: Calculate final concentrations

The initial concentration of COCl2 = 1.6M

The initial concentration of CO and Cl2 = 0M

There will react xM of COCl2

Since the mole ratio is 1:1

The final concentration of CO and Cl2 will be X M

The final concentration of COCl2 will be (1.6 -X)M

Step 4: Define Kc

Kc= [CO] *[Cl2] / [COCl2] = 8.33*10^-4

Kc = X*X / 1.6-X = 8.33 * 10^-4

8.33 * 10^-4 = X² /(1.6-X)

8.33 * 10^-4 *(1.6 -X) = X²

0.0013328 - 8.33*10^-4 X = X²

X² + 8.33*10^-4 X - 0.0013328= 0

X = 0.0361 M = [CO] = [Cl2]

[COCl2] = 1.6 - 0.0361 = 1.5639 M

To control this we can calculate the Kc

(0.0361*0.0361)/1.5639 = 0.000833

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A solution was prepared by dissolving 195.0 g of KCl in 215 g of water. Calculate the mole fraction of KCl. (The formula weight

defon

Answer:

Approximately $0.180$ .

Explanation:

The mole fraction of a compound in a solution is:

$\displaystyle \frac{\text{Number of moles of compound in question}}{\text{Number of moles of all particles in the solution}}$ .

In this question, the mole fraction of $\rm KCl$ in this solution would be:

$\displaystyle X_\mathrm{KCl} = \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}$ .

This solution consist of only $\rm KCl$ and water (i.e., $\rm H_2O$ .) Hence:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})}\end{aligned}$ .

From the question:

Mass of $\rm KCl$ : $m(\mathrm{KCl}) = 195.0\; \rm g$ .
Molar mass of $\rm KCl$ : $M(\mathrm{KCl}) = 74.6\; \rm g \cdot mol^{-1}$ .

Mass of $\rm H_2O$ : $m(\mathrm{H_2O}) = 215\; \rm g$ .
Molar mass of $\rm H_2O$ : $M(\mathrm{H_2O}) = 18.0\; \rm g\cdot mol^{-1}$ .

Apply the formula $\displaystyle n = \frac{m}{M}$ to find the number of moles of $\rm KCl$ and $\rm H_2O$ in this solution.

$\begin{aligned}n(\mathrm{KCl}) &= \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} \\ &= \frac{195.0\; \rm g}{74.6\; \rm g \cdot mol^{-1}} \approx 2.61\; \em \rm mol\end{aligned}$ .

$\begin{aligned}n(\mathrm{H_2O}) &= \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} \\ &= \frac{215\; \rm g}{18.0\; \rm g \cdot mol^{-1}} \approx 11.9\; \em \rm mol\end{aligned}$ .

The molar fraction of $\rm KCl$ in this solution would be:

$\begin{aligned} X_\mathrm{KCl} &= \frac{n(\mathrm{KCl})}{n(\text{All particles in this soluton})}\\ &= \frac{n(\mathrm{KCl})}{n(\mathrm{KCl}) + n(\mathrm{H_2O})} \\ &\approx \frac{2.61 \; \rm mol}{2.61\; \rm mol + 11.9\; \rm mol} \approx 0.180\end{aligned}$ .

(Rounded to three significant figures.)

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