Nitrogen and hydrogen react to form ammonia, like this: N2(9) + 3 H2(g) → 2 NH3(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of nitrogen, hydrogen, and ammonia has the following composition: compound concentration at equilibrium N2 1.4M H2 0.81M NH3 1.9M Calculate the value of the equilibrium constant K, for this reaction. Round your answer to 2 significant digits. K_ = 0
Nitrogen and hydrogen react to form ammonia, like this: N2(9) + 3 H2(g) → 2 NH3(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of nitrogen, hydrogen, and ammonia has the following composition: compound concentration at equilibrium N2 1.4M H2 0.81M NH3 1.9M Calculate the value of the equilibrium constant K, for this reaction. Round your answer to 2 significant digits. K_ = 0
Chemistry
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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
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![**Nitrogen and Hydrogen Reaction to Form Ammonia**
In this reaction, nitrogen and hydrogen react to form ammonia, following the chemical equation:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
At a certain temperature, a chemist finds that the equilibrium mixture of nitrogen, hydrogen, and ammonia has the following composition:
| Compound | Concentration at Equilibrium |
|----------|------------------------------|
| \(\text{N}_2\) | 1.4 M |
| \(\text{H}_2\) | 0.81 M |
| \(\text{NH}_3\) | 1.9 M |
To calculate the value of the equilibrium constant \( K_c \) for this reaction, use the provided concentrations. Ensure your answer is rounded to 2 significant digits.
The table above summarizes the equilibrium concentrations of nitrogen (\(\text{N}_2\)), hydrogen (\(\text{H}_2\)), and ammonia (\(\text{NH}_3\)) in molarity (M). This information is essential for determining the equilibrium constant.
**Calculation:**
\[ K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Using the equilibrium concentrations:
\[ [\text{N}_2] = 1.4 \, M \]
\[ [\text{H}_2] = 0.81 \, M \]
\[ [\text{NH}_3] = 1.9 \, M \]
Substitute these values into the equation:
\[ K_c = \dfrac{(1.9)^2}{(1.4) \cdot (0.81)^3} \]
After performing the calculation, ensure you round \( K_c \) to 2 significant digits.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F90b48bfa-6798-4a35-8945-d86c1a55346c%2Ffc1fefec-d908-459a-adef-b523dc5eed91%2Fwahmi4u_processed.png&w=3840&q=75)
Transcribed Image Text:**Nitrogen and Hydrogen Reaction to Form Ammonia**
In this reaction, nitrogen and hydrogen react to form ammonia, following the chemical equation:
\[ \text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) \]
At a certain temperature, a chemist finds that the equilibrium mixture of nitrogen, hydrogen, and ammonia has the following composition:
| Compound | Concentration at Equilibrium |
|----------|------------------------------|
| \(\text{N}_2\) | 1.4 M |
| \(\text{H}_2\) | 0.81 M |
| \(\text{NH}_3\) | 1.9 M |
To calculate the value of the equilibrium constant \( K_c \) for this reaction, use the provided concentrations. Ensure your answer is rounded to 2 significant digits.
The table above summarizes the equilibrium concentrations of nitrogen (\(\text{N}_2\)), hydrogen (\(\text{H}_2\)), and ammonia (\(\text{NH}_3\)) in molarity (M). This information is essential for determining the equilibrium constant.
**Calculation:**
\[ K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \]
Using the equilibrium concentrations:
\[ [\text{N}_2] = 1.4 \, M \]
\[ [\text{H}_2] = 0.81 \, M \]
\[ [\text{NH}_3] = 1.9 \, M \]
Substitute these values into the equation:
\[ K_c = \dfrac{(1.9)^2}{(1.4) \cdot (0.81)^3} \]
After performing the calculation, ensure you round \( K_c \) to 2 significant digits.
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