Question4. \mathrm{HP} \AA \grave{A} une certaine température, $K_{\mathrm{c}}$ égale 4,8 dans la réaction suivante entre le dioxyde de soufre, $\mathrm{SO}_{2}(\mathrm{~g})$ , et le dioxyde d'azote, $\mathrm{NO}_{2}(\mathrm{~g})$ : $\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}(\mathrm{~g})+\mathrm{SO}_{3}(\mathrm{~g})$
Si le dioxyde de soufre et le dioxyde d'azote ont une concentration initiale de $0,36 \mathrm{~mol} / \mathrm{L}$ , quelle quantité de trioxyde de soufre, $\mathrm{SO}_{3}(\mathrm{~g})$ , est présente dans un récipient de 5,0 L à l'état d'équilibre?

Studdy Solution

STEP 1

1. The reaction is at equilibrium with a given equilibrium constant $K_c = 4.8$ .
2. The initial concentrations of $\mathrm{SO}_2$ and $\mathrm{NO}_2$ are both $0.36 \, \mathrm{mol/L}$ .
3. The volume of the container is $5.0 \, \mathrm{L}$ .
4. We need to find the equilibrium concentration of $\mathrm{SO}_3$ .

STEP 2

1. Set up the equilibrium expression for the reaction.
2. Use an ICE (Initial, Change, Equilibrium) table to express concentrations at equilibrium.
3. Solve for the equilibrium concentration of $\mathrm{SO}_3$ .
4. Calculate the amount of $\mathrm{SO}_3$ in moles.

STEP 3

The equilibrium expression for the reaction is given by:
$K_c = \frac{[\mathrm{NO}][\mathrm{SO}_3]}{[\mathrm{SO}_2][\mathrm{NO}_2]}$

STEP 4

Set up an ICE table:
- Initial concentrations: $[\mathrm{SO}_2] = 0.36 \, \mathrm{mol/L}$ , $[\mathrm{NO}_2] = 0.36 \, \mathrm{mol/L}$ , $[\mathrm{NO}] = 0 \, \mathrm{mol/L}$ , $[\mathrm{SO}_3] = 0 \, \mathrm{mol/L}$ . - Change in concentrations: $-x$ for $[\mathrm{SO}_2]$ and $[\mathrm{NO}_2]$ , $+x$ for $[\mathrm{NO}]$ and $[\mathrm{SO}_3]$ . - Equilibrium concentrations: $[\mathrm{SO}_2] = 0.36 - x$ , $[\mathrm{NO}_2] = 0.36 - x$ , $[\mathrm{NO}] = x$ , $[\mathrm{SO}_3] = x$ .

STEP 5

Substitute the equilibrium concentrations into the equilibrium expression:
$K_c = \frac{x \cdot x}{(0.36 - x)(0.36 - x)} = 4.8$
Simplify and solve for $x$ :
$\frac{x^2}{(0.36 - x)^2} = 4.8$
Take the square root of both sides:
$\frac{x}{0.36 - x} = \sqrt{4.8}$
Solve for $x$ :
$x = 0.36 \cdot \frac{\sqrt{4.8}}{1 + \sqrt{4.8}}$

STEP 6

Calculate the amount of $\mathrm{SO}_3$ in moles:
$x = [\mathrm{SO}_3] = 0.36 \cdot \frac{\sqrt{4.8}}{1 + \sqrt{4.8}}$
$\text{Amount of } \mathrm{SO}_3 = x \times 5.0 \, \mathrm{L}$
Calculate the numerical value to find the amount in moles.
The amount of $\mathrm{SO}_3$ in the container at equilibrium is approximately:
$\boxed{0.72 \, \text{moles}}$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms