### Chemical Equilibrium CalculationAt a certain temperature, 0.312 mol of CH₄ and 0.728 mol of H₂S are placed in a 2.00 L container.The chemical reaction is:\[ \text{CH}_4(\text{g}) + 2\text{H}_2\text{S}(\text{g}) \rightleftharpoons \text{CS}_2(\text{g}) + 4\text{H}_2(\text{g}) \]At equilibrium, 7.92 g of CS₂ is present. Calculate \( K_c \).To begin, we need to convert the mass of CS₂ to moles:\[ \text{Moles of } \text{CS}_2 = \frac{7.92 \text{ g}}{\text{Molar Mass of } \text{CS}_2} \]The molar mass of CS₂ is:\[ \text{Molar Mass of } \text{CS}_2 = 12.01 \text{ (Carbon)} + 2 \times 32.07 \text{ (Sulfur)} = 76.15 \text{ g/mol} \]So, the moles of CS₂ are:\[ \frac{7.92 \text{ g}}{76.15 \text{ g/mol}} \approx 0.104 \text{ mol} \]Next, let's set up an I.C.E. (Initial, Change, Equilibrium) table:| | CH₄ (g) | H₂S (g) | CS₂ (g) | H₂ (g) ||---------------|-----------|-----------|-----------|-----------|| Initial (mol) | 0.312 | 0.728 | 0 | 0 || Change (mol) | -x | -2x | +x | +4x || Equilibrium | 0.312 - x | 0.728 - 2x| x | 4x |From above, we have the equilibrium concentration of CS₂:\[ x = 0.104 \text{ mol} \]Using the change \( x \), we find:\[ \text{Equilibrium concentration of CH}_4 = 0.312 - 0.104 =

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At a certain temperature, 0.312 mol CH, and 0.728 mol H, S are placed in a 2.00 L container. CH(g) + 2 H, S(g) = CS,(g) + 4 H, (g) At equilibrium, 7.92 g CS, is present. Calculate K.. K. = 0.11 Incorrect

At a certain temperature, 0.312 mol CH, and 0.728 mol H, S are placed in a 2.00 L container. CH(g) + 2 H, S(g) = CS,(g) + 4 H, (g) At equilibrium, 7.92 g CS, is present. Calculate K.. K. = 0.11 Incorrect

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

### Chemical Equilibrium Calculation

At a certain temperature, 0.312 mol of CH₄ and 0.728 mol of H₂S are placed in a 2.00 L container.

The chemical reaction is:
\[ \text{CH}_4(\text{g}) + 2\text{H}_2\text{S}(\text{g}) \rightleftharpoons \text{CS}_2(\text{g}) + 4\text{H}_2(\text{g}) \]

At equilibrium, 7.92 g of CS₂ is present. Calculate \( K_c \).

To begin, we need to convert the mass of CS₂ to moles:
\[ \text{Moles of } \text{CS}_2 = \frac{7.92 \text{ g}}{\text{Molar Mass of } \text{CS}_2} \]

The molar mass of CS₂ is:
\[ \text{Molar Mass of } \text{CS}_2 = 12.01 \text{ (Carbon)} + 2 \times 32.07 \text{ (Sulfur)} = 76.15 \text{ g/mol} \]

So, the moles of CS₂ are:
\[ \frac{7.92 \text{ g}}{76.15 \text{ g/mol}} \approx 0.104 \text{ mol} \]

Next, let's set up an I.C.E. (Initial, Change, Equilibrium) table:

| | CH₄ (g) | H₂S (g) | CS₂ (g) | H₂ (g) |
|---------------|-----------|-----------|-----------|-----------|
| Initial (mol) | 0.312 | 0.728 | 0 | 0 |
| Change (mol) | -x | -2x | +x | +4x |
| Equilibrium | 0.312 - x | 0.728 - 2x| x | 4x |

From above, we have the equilibrium concentration of CS₂:
\[ x = 0.104 \text{ mol} \]

Using the change \( x \), we find:
\[ \text{Equilibrium concentration of CH}_4 = 0.312 - 0.104 =

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