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3 years ago

The equilibrium constant Kc for the reaction

Chemistry

1 answer:

PIT_PIT [208]3 years ago

3 0

Answer:

[N2] = 0.3633M

[H2] = 1.090M

[NH3] = 0.2734M

Explanation:

Based on the reaction of the problem, Kc is defined as:

Kc = 0.159 = [NH3]² / [N2] [H2]³

<em>Where [] are the equilibrium concentrations.</em>

The initial concentrations of the reactants is:

N2 = 1.00mol / 2.00L = 0.500M

H2 = 3.00mol / 2.00L = 1.50M

When the equilibrium is reached, the concentrations are:

[N2] = 0.500M - X

[H2] = 1.50M - 3X

[NH3] = 2X

<em>Where X is reaction quotient</em>

Replacing in the Kc equation:

0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³

0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴

0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²

0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0

Solving for X:

X = 0.1367. Right solution.

X = 1.8286. False solution. Produce negative concentrations

Replacing:

[N2] = 0.500M - 0.1367M

[H2] = 1.50M - 3*0.1367M

[NH3] = 2*0.1367M

The equilibrium concentrations are:

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Step 1: Given data

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stich3 [128]

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Given data

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<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

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