Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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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Transcribed Image Text:### 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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