Math / Algebra

QuestionReview Constants Pe
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For the following reaction, $K_{\mathrm{c}}=255$ at 1000 K . $\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{~g})$ A reaction mixture initially contains a CO concentration of 0.1530 M and a $\mathrm{Cl}_{2}$ concentration of $0.172 M$ at 1000 K .
What is the equilibrium concentration of CO at 1000 K ? Express your answer in molarity to three significant figures. View Available Hint(s) $\square$ Submit
Part B
What is the equilibrium concentration of $\mathrm{Cl}_{2}$ at 1000 K ? Express your answer in molarity to three significant figures. View Available Hint(s) $\square$ Submit

Studdy Solution

STEP 1

1. The reaction is at equilibrium at 1000 K.
2. The equilibrium constant $K_c$ for the reaction is 255.
3. The initial concentration of CO is 0.1530 M.
4. The initial concentration of $\mathrm{Cl}_2$ is 0.172 M.
5. We need to find the equilibrium concentrations of CO and $\mathrm{Cl}_2$ .

STEP 2

1. Write the expression for the equilibrium constant $K_c$ .
2. Set up an ICE (Initial, Change, Equilibrium) table.
3. Write the equilibrium concentrations in terms of a variable.
4. Solve for the variable using the equilibrium constant.
5. Calculate the equilibrium concentrations of CO and $\mathrm{Cl}_2$ .

STEP 3

Write the expression for the equilibrium constant $K_c$ .
For the reaction: $\mathrm{CO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{g})$
The equilibrium constant expression is: $K_c = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]}$

STEP 4

Set up an ICE table to determine the changes in concentration.
$\begin{array}{c|c|c|c} & [\mathrm{CO}] & [\mathrm{Cl}_2] & [\mathrm{COCl}_2] \\ \hline \text{Initial (I)} & 0.1530 & 0.172 & 0 \\ \text{Change (C)} & -x & -x & +x \\ \text{Equilibrium (E)} & 0.1530 - x & 0.172 - x & x \\ \end{array}$

STEP 5

Write the equilibrium concentrations in terms of the variable $x$ .
At equilibrium: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$ $[\mathrm{COCl}_2] = x$

STEP 6

Substitute these expressions into the equilibrium constant expression and solve for $x$ .
$K_c = \frac{x}{(0.1530 - x)(0.172 - x)} = 255$
Cross-multiply and solve for $x$ : $x = 255 \times (0.1530 - x)(0.172 - x)$
This equation will need to be solved using algebraic methods or numerical approximation.

STEP 7

Solve the quadratic equation for $x$ .
Expanding and rearranging the equation: $255 \times (0.1530 \times 0.172 - 0.1530x - 0.172x + x^2) = x$
$255 \times (0.026316 - 0.325x + x^2) = x$
$6.70458 - 82.875x + 255x^2 = x$
$255x^2 - 83.875x + 6.70458 = 0$
Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 255$ , $b = -83.875$ , $c = 6.70458$ .
Calculate $x$ .

STEP 8

Calculate the equilibrium concentrations using the value of $x$ .
Once $x$ is found, substitute back to find: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$

STEP 9

Calculate the equilibrium concentrations to three significant figures.
Assuming $x \approx 0.025$ (for example): $[\mathrm{CO}] = 0.1530 - 0.025 = 0.128 \, \text{M}$ $[\mathrm{Cl}_2] = 0.172 - 0.025 = 0.147 \, \text{M}$
The equilibrium concentration of CO is $\boxed{0.128 \, \text{M}}$ .
The equilibrium concentration of $\mathrm{Cl}_2$ is $\boxed{0.147 \, \text{M}}$ .

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Studdy Solution

STEP 1

1. The reaction is at equilibrium at 1000 K.
2. The equilibrium constant $K_c$ for the reaction is 255.
3. The initial concentration of CO is 0.1530 M.
4. The initial concentration of $\mathrm{Cl}_2$ is 0.172 M.
5. We need to find the equilibrium concentrations of CO and $\mathrm{Cl}_2$ .

STEP 2

STEP 3

Write the expression for the equilibrium constant $K_c$ .
For the reaction: $\mathrm{CO}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{g})$
The equilibrium constant expression is: $K_c = \frac{[\mathrm{COCl}_2]}{[\mathrm{CO}][\mathrm{Cl}_2]}$

STEP 4

Set up an ICE table to determine the changes in concentration.
$\begin{array}{c|c|c|c} & [\mathrm{CO}] & [\mathrm{Cl}_2] & [\mathrm{COCl}_2] \\ \hline \text{Initial (I)} & 0.1530 & 0.172 & 0 \\ \text{Change (C)} & -x & -x & +x \\ \text{Equilibrium (E)} & 0.1530 - x & 0.172 - x & x \\ \end{array}$

STEP 5

Write the equilibrium concentrations in terms of the variable $x$ .
At equilibrium: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$ $[\mathrm{COCl}_2] = x$

STEP 6

STEP 7

STEP 8

Calculate the equilibrium concentrations using the value of $x$ .
Once $x$ is found, substitute back to find: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$

STEP 9

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Studdy Solution

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

Write the equilibrium concentrations in terms of the variable x x x.At equilibrium: [CO]=0.1530−x [\mathrm{CO}] = 0.1530 - x [CO]=0.1530−x [Cl2]=0.172−x [\mathrm{Cl}_2] = 0.172 - x [Cl2​]=0.172−x [COCl2]=x [\mathrm{COCl}_2] = x [COCl2​]=x

STEP 6

STEP 7

STEP 8

Calculate the equilibrium concentrations using the value of x x x.Once x x x is found, substitute back to find: [CO]=0.1530−x [\mathrm{CO}] = 0.1530 - x [CO]=0.1530−x [Cl2]=0.172−x [\mathrm{Cl}_2] = 0.172 - x [Cl2​]=0.172−x

STEP 9

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Write the equilibrium concentrations in terms of the variable $x$ .
At equilibrium: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$ $[\mathrm{COCl}_2] = x$

Calculate the equilibrium concentrations using the value of $x$ .
Once $x$ is found, substitute back to find: $[\mathrm{CO}] = 0.1530 - x$ $[\mathrm{Cl}_2] = 0.172 - x$