### Chemical Equilibrium ProblemAn initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction:\[ 3 \text{H}_2(\text{g}) + \text{N}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g}) \]At equilibrium, the concentrations are $[\text{H}_2] = 6.0 \, \text{M}$, $[\text{N}_2] = 12 \, \text{M}$, and $[\text{NH}_3] = 4.0 \, \text{M}$. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?- $[\text{H}_2]_0 =$ [Input field] M- $[\text{N}_2]_0 =$ [Input field] M **Chemical Equilibrium Problem:**Consider the following system at equilibrium where $ \Delta H^\circ = 92.7 \, \text{kJ} $, and $ K_c = 1.80 \times 10^{-4} $, at 298 K:\[ \text{NH}_4\text{HS (s)} \rightarrow \text{NH}_3\text{ (g)} + \text{H}_2\text{S (g)} \]**Scenario:**When some moles of NH$_4$HS(s) are removed from the equilibrium system at constant temperature:1. **The value of $ K_c $:** - $ \circ $ Increases - $ \bigcirc $ Decreases - $ \circ $ Remains the same2. **The value of $ Q_c $:** - $ \circ $ Is greater than $ K_c $ - $ \circ $ Is equal to $ K_c $ - $ \bigcirc $ Is less than $ K_c $3. **The reaction must:** - $ \bigcirc $ Run in the forward direction to reestablish equilibrium. - $ \circ $ Run in the reverse direction to reestablish equilibrium. - $ \circ $ Remain the same. It is already at equilibrium.4. **The number of moles of NH$_3$ will:** - $ \bigcirc $ Increase - $ \circ $ Decrease - $ \circ $ Remain the same**Explanation:**- **Equilibrium Constant $ K_c $:** The equilibrium constant $ K_c $ remains unchanged with the removal of a reactant or product as it solely depends on temperature.- **Reaction Quotient $ Q_c $:** Removing the solid NH$_4$HS shifts the balance by reducing the concentration of products. Hence, $ Q_c $ becomes less than $ K_c $, making the system shift towards products to restore equilibrium.- **Reaction Direction:** With $ Q_c $ less than $ K_c $, the reaction shifts forward to form more products and restore equilibrium.- **NH$_3$ Concentration:** As the reaction proceeds forward, the concentration of NH\

Answered: An initial mixture of nitrogen gas and…

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...

See similar textbooks

Related questions

Question

### Chemical Equilibrium Problem

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction:

\[
3 \text{H}_2(\text{g}) + \text{N}_2(\text{g}) \rightleftharpoons 2 \text{NH}_3(\text{g})
\]

At equilibrium, the concentrations are $[\text{H}_2] = 6.0 \, \text{M}$, $[\text{N}_2] = 12 \, \text{M}$, and $[\text{NH}_3] = 4.0 \, \text{M}$. What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

- $[\text{H}_2]_0 =$ [Input field] M
- $[\text{N}_2]_0 =$ [Input field] M

**Chemical Equilibrium Problem:**

Consider the following system at equilibrium where $ \Delta H^\circ = 92.7 \, \text{kJ} $, and $ K_c = 1.80 \times 10^{-4} $, at 298 K:

\[ \text{NH}_4\text{HS (s)} \rightarrow \text{NH}_3\text{ (g)} + \text{H}_2\text{S (g)} \]

**Scenario:**
When some moles of NH$_4$HS(s) are removed from the equilibrium system at constant temperature:

1. **The value of $ K_c $:**
- $ \circ $ Increases
- $ \bigcirc $ Decreases
- $ \circ $ Remains the same

2. **The value of $ Q_c $:**
- $ \circ $ Is greater than $ K_c $
- $ \circ $ Is equal to $ K_c $
- $ \bigcirc $ Is less than $ K_c $

3. **The reaction must:**
- $ \bigcirc $ Run in the forward direction to reestablish equilibrium.
- $ \circ $ Run in the reverse direction to reestablish equilibrium.
- $ \circ $ Remain the same. It is already at equilibrium.

4. **The number of moles of NH$_3$ will:**
- $ \bigcirc $ Increase
- $ \circ $ Decrease
- $ \circ $ Remain the same

**Explanation:**

- **Equilibrium Constant $ K_c $:**
The equilibrium constant $ K_c $ remains unchanged with the removal of a reactant or product as it solely depends on temperature.

- **Reaction Quotient $ Q_c $:**
Removing the solid NH$_4$HS shifts the balance by reducing the concentration of products. Hence, $ Q_c $ becomes less than $ K_c $, making the system shift towards products to restore equilibrium.

- **Reaction Direction:**
With $ Q_c $ less than $ K_c $, the reaction shifts forward to form more products and restore equilibrium.

- **NH$_3$ Concentration:**
As the reaction proceeds forward, the concentration of NH\

Expert Solution

Step 1: Given data

“Since you have posted multiple questions, we will provide the solution only to the first question as per our Q&A guidelines. Please repost the remaining questions separately.”

Given:

$3 space straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight N subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space 2 space NH subscript 3 left parenthesis straight g right parenthesis$

From the equilibrium concentrations are:
[H₂] = 6.0 M
[N₂] = 12 M
[NH₃] = 4.0 M

Chemical equilibrium is a dynamic state in a reversible chemical reaction where the rates of the forward and reverse reactions become equal. This means that while reactants are still forming products and products are forming reactants, their rates of conversion are balanced. At equilibrium, the concentrations of products and reactants generally stabilize, but they are not necessarily identical.

Step by step

Solved in 6 steps with 12 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions