Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S,(g)+C(s) = CS,(g) K. = 9.40 at 900 K How many grams of CS,(g) can be prepared by heating 13.1 mol S,(g) with excess carbon in a 8.50 L reaction vessel held at 900 K until equilibrium is attained? mass of CS,(g):

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**General Chemistry 4th Edition**
*McQuarrie • Rock • Gallogly*

**University Science Books**  
presented by Macmillan Learning

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Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is:

\[ 
\text{S}_2(g) + \text{C}(s) \rightleftharpoons \text{CS}_2(g) \quad K_c = 9.40 \, \text{at} \, 900 \, \text{K}
\]

How many grams of \(\text{CS}_2(g)\) can be prepared by heating 13.1 mol \(\text{S}_2(g)\) with excess carbon in an 8.50 L reaction vessel held at 900 K until equilibrium is attained?

**mass of \(\text{CS}_2(g)\):** \_\_\_\_\_ g

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Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry | Publisher: University Science Books

*Note: This is a text-based transcription of an educational content image for learning purposes.*
Transcribed Image Text:**General Chemistry 4th Edition** *McQuarrie • Rock • Gallogly* **University Science Books** presented by Macmillan Learning --- Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is: \[ \text{S}_2(g) + \text{C}(s) \rightleftharpoons \text{CS}_2(g) \quad K_c = 9.40 \, \text{at} \, 900 \, \text{K} \] How many grams of \(\text{CS}_2(g)\) can be prepared by heating 13.1 mol \(\text{S}_2(g)\) with excess carbon in an 8.50 L reaction vessel held at 900 K until equilibrium is attained? **mass of \(\text{CS}_2(g)\):** \_\_\_\_\_ g --- Question Source: McQuarrie, Rock, and Gallogly 4e - General Chemistry | Publisher: University Science Books *Note: This is a text-based transcription of an educational content image for learning purposes.*
**Equilibrium Concentration Calculation**

**Consider the reaction:**

\[ \text{A(aq)} \rightleftharpoons 2\text{B(aq)} \]

\[ K_c = 4.46 \times 10^{-6} \text{ at 500 K} \]

**Problem Statement:**

If a 4.70 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?

\[ \text{[B]} = \_\_\_\_\_\_\_\_ \text{ M} \]

**Explanation:**

This is a classic equilibrium problem where you are given the initial concentration of A and the equilibrium constant (\(K_c\)) at a particular temperature (500 K). You are asked to find the equilibrium concentration of product B. 

To solve this:

1. **Set up an ICE table (Initial, Change, Equilibrium) for the reaction.**

2. **Use the expression for the equilibrium constant:**

   \[ K_c = \frac{[\text{B}]^2}{[\text{A}]} \]

   **where \([\text{A}]\) is the concentration of A at equilibrium, and \([\text{B}]\) is the concentration of B at equilibrium.**

3. **Solve for \([\text{B}]\) using the given values and the assumed changes at equilibrium.**

This approach helps in visually organizing how the concentrations change and applying the stoichiometry of the reaction to find the solution.
Transcribed Image Text:**Equilibrium Concentration Calculation** **Consider the reaction:** \[ \text{A(aq)} \rightleftharpoons 2\text{B(aq)} \] \[ K_c = 4.46 \times 10^{-6} \text{ at 500 K} \] **Problem Statement:** If a 4.70 M sample of A is heated to 500 K, what is the concentration of B at equilibrium? \[ \text{[B]} = \_\_\_\_\_\_\_\_ \text{ M} \] **Explanation:** This is a classic equilibrium problem where you are given the initial concentration of A and the equilibrium constant (\(K_c\)) at a particular temperature (500 K). You are asked to find the equilibrium concentration of product B. To solve this: 1. **Set up an ICE table (Initial, Change, Equilibrium) for the reaction.** 2. **Use the expression for the equilibrium constant:** \[ K_c = \frac{[\text{B}]^2}{[\text{A}]} \] **where \([\text{A}]\) is the concentration of A at equilibrium, and \([\text{B}]\) is the concentration of B at equilibrium.** 3. **Solve for \([\text{B}]\) using the given values and the assumed changes at equilibrium.** This approach helps in visually organizing how the concentrations change and applying the stoichiometry of the reaction to find the solution.
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