Problem 90
Question
Jupiter's Atmosphere Ammonium hydrogen sulfide \(\left(\mathrm{NH}_{4} \mathrm{SH}\right)\) has been detected in the atmosphere of Jupiter, where it probably exists in equilibrium with ammonia and hydrogen sulfide: $$ \mathrm{NH}_{1} \mathrm{SH}(j) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) $$ The value of \(K_{p}\) for the reaction at \(24^{\circ} \mathrm{C}\) is \(0.126 .\) Suppose a sealed flask contains an equilibrium mixture of \(\mathrm{NH}_{4} \mathrm{SH}\), \(\mathrm{NH}_{3},\) and \(\mathrm{H}_{2} \mathrm{S}\). At equilibrium, the partial pressure of \(\mathrm{H}_{2} \mathrm{S}\) is 0.355 atm. What is the partial pressure of \(\mathrm{NH}_{3} ?\)
Step-by-Step Solution
Verified Answer
Answer: The partial pressure of ammonia (NH3) at equilibrium is approximately 0.355 atm.
1Step 1: Write the reaction and the expression for Kp
We have the reaction:
$$
\mathrm{NH}_{4} \mathrm{SH}(j) \rightleftharpoons
\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)
$$
And the expression for Kp is:
$$
K_p = \frac{P_{NH_3} \times P_{H_2S}}{P_{NH_4SH}}
$$
2Step 2: Find the pressure of NH3 at equilibrium
We are given the equilibrium constant, Kp, value as \(0.126\) and the partial pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium as \(0.355\ \mathrm{atm}\).
Now, let the partial pressure of \(\mathrm{NH}_{3}\) be \(P_{NH_3}\). Since \(\mathrm{NH}_{4} \mathrm{SH}\) does not contribute to the pressure, we can assume its partial pressure as 1. Then, the expression for Kp becomes:
$$
0.126 = \frac{P_{NH_3} \times 0.355}{1}
$$
3Step 3: Calculate the partial pressure of NH3
To find the partial pressure of NH3, we solve for \(P_{NH_3}\):
$$
P_{NH_3} = \frac{0.126}{0.355}
$$
4Step 4: Determine the partial pressure of NH3
Now, calculate the partial pressure:
$$
P_{NH_3} = \frac{0.126}{0.355} \approx 0.355 \mathrm{atm}
$$
At equilibrium, the partial pressure of ammonia (NH3) is approximately \(0.355\ \mathrm{atm}\).
Key Concepts
Ammonium Hydrogen SulfidePartial PressureJupiter's AtmosphereEquilibrium Constant
Ammonium Hydrogen Sulfide
Ammonium hydrogen sulfide, often written as \( \text{NH}_4\text{SH} \), is a compound that plays an interesting role in chemistry, and specifically in the context of Jupiter's atmosphere. It is a combination of ammonia (\( \text{NH}_3 \)) and hydrogen sulfide (\( \text{H}_2\text{S} \)). This compound is significant for several reasons:
- It can decompose into ammonia and hydrogen sulfide, which are both gases at certain conditions.
- This decomposition reaction is central to the equilibrium process discussed in Jupiter’s atmosphere.
- The study of \( \text{NH}_4\text{SH} \) and its reactions help researchers understand atmospheric processes and conditions.
Partial Pressure
The concept of partial pressure is crucial in understanding gas mixtures. Each gas in a mixture contributes to the total pressure exerted by the gases. In our context, it refers to:
- The pressure that ammonia (\( \text{NH}_3 \)) exerts alone in the mixture with other gases.
- The pressure that hydrogen sulfide (\( \text{H}_2\text{S} \)) exerts alone in the mixture.
Jupiter's Atmosphere
Exploring Jupiter's atmosphere reveals a fascinating mixture of gases, different from Earth's. The atmosphere contains compounds such as ammonium hydrogen sulfide, ammonia, and hydrogen sulfide.
- This atmosphere is mainly composed of hydrogen and helium, but these additional compounds add complexity to its composition.
- The behavior of ammonium hydrogen sulfide and its decomposition involving ammonia and hydrogen sulfide provides insight into the atmospheric chemistry.
Equilibrium Constant
The equilibrium constant, symbolized as \( K_p \) when dealing with gases, is a fundamental aspect of chemical reactions at equilibrium. It provides a numerical value that helps predict the proportions of reactants and products at equilibrium:
- For the decomposition of ammonium hydrogen sulfide in the atmosphere, \( K_p \) represents the ratio of the partial pressures of ammonia and hydrogen sulfide.
- The given \( K_p \) value of 0.126 is used to determine the equilibrium state under the specific conditions provided.
Other exercises in this chapter
Problem 86
The value of \(K_{p}\) for the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.3 \times 10^{-4}\) at \(648 \
View solution Problem 88
Making Hydrogen Gas Passing steam over hot carbon produces a mixture of carbon monoxide and hydrogen: $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{C}(s) \rightleftha
View solution Problem 91
A flask containing pure \(\mathrm{NO}_{2}\) is heated to \(1000 \mathrm{K},\) a temperature at which \(K_{p}=158\) for the decomposition of \(\mathrm{NO}_{2}\)
View solution Problem 92
The equilibrium constant \(K_{\mathrm{p}}\) of the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ is 7.69 at \(83
View solution