An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperatur

Q: Why don't we include the solid $\text{NH}_4\text{HS}$ in the $K_p$ expression?

The concentration or active mass of a pure solid or pure liquid is constant during a reaction, regardless of its amount. Therefore, it is incorporated into the equilibrium constant itself and is conventionally taken as $1$.

Q: Why is the initial pressure of $\text{H}_2\text{S}$ zero?

The problem states the flask *already* contained ammonia gas before the decomposition started. It doesn't mention any initial $\text{H}_2\text{S}$, so we assume it's zero until the solid begins to decompose.

Q: Can we use concentrations ($K_c$) instead of partial pressures ($K_p$)?

You could, but it would be an unnecessary extra step. Since all the given data (initial state and equilibrium state) are in terms of pressure (atm), it is much more direct and efficient to calculate $K_p$ directly.

Q: What if the volume of the flask was given? Would it change the answer?

No. The equilibrium constant $K_p$ depends only on temperature. Changing the volume would shift the equilibrium position (changing the number of moles of gas to maintain $K_p$), but the value of $K_p$ itself would remain exactly the same.

Q: Why did we add $x$ to both gases?

Look at the stoichiometry of the balanced equation: $\text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g)$. For every $1$ mole of solid that decomposes, exactly $1$ mole of $\text{NH}_3$ and $1$ mole of $\text{H}_2\text{S}$ are produced. Therefore, their pressures must increase by the exact same amount, $x$.

Question

An amount of solid $	ext{NH}_4	ext{HS}$ is placed in a flask already containing ammonia gas at a certain temperature and $0.50	ext{ atm}$ pressure. Ammonium hydrogen sulphide decomposes to yield $	ext{NH}_3$ and $	ext{H}_2	ext{S}$ gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to $0.84	ext{ atm}$. The equilibrium constant for the decomposition of $	ext{NH}_4	ext{HS}$ at this temperature is

Options:

(A) $0.11$

(B) $0.17$

(C) $0.18$

(D) $0.30$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a sealed flask. Inside, we have a chunk of solid ammonium hydrogen sulfide ($	ext{NH}_4	ext{HS}$). But the flask isn't empty; it already contains ammonia ($	ext{NH}_3$) gas exerting a pressure of $0.50 	ext{ atm}$. This is our initial state. The presence of this initial ammonia is a classic twist in chemical equilibrium problems, often referred to as the "common ion effect" in aqueous solutions, but here it's a "common gas effect".

### The Master Equation

When the solid $	ext{NH}_4	ext{HS}$ begins to decompose, it establishes the following equilibrium:

$$ 	ext{NH}_4	ext{HS}(s) ightleftharpoons 	ext{NH}_3(g) + 	ext{H}_2	ext{S}(g) $$

Notice that for every mole of solid that decomposes, it produces one mole of $	ext{NH}_3$ and one mole of $	ext{H}_2	ext{S}$. Let's assume that at equilibrium, the decomposition has added $x 	ext{ atm}$ to the pressure of each gas.

We can set up an ICE (Initial, Change, Equilibrium) table to track the pressures:
- **Initial:** $P_{	ext{NH}_3} = 0.50 	ext{ atm}$, $P_{	ext{H}_2	ext{S}} = 0 	ext{ atm}$
- **Change:** $P_{	ext{NH}_3}$ increases by $x$, $P_{	ext{H}_2	ext{S}}$ increases by $x$
- **Equilibrium:** $P_{	ext{NH}_3} = (0.50 + x) 	ext{ atm}$, $P_{	ext{H}_2	ext{S}} = x 	ext{ atm}$

### The Pressure Constraint

The problem gives us a crucial piece of information: the total pressure at equilibrium is $0.84 	ext{ atm}$. According to Dalton's Law of Partial Pressures, the total pressure is simply the sum of the individual partial pressures of the gases present.

$$ P_{	ext{total}} = P_{	ext{NH}_3} + P_{	ext{H}_2	ext{S}} $$

Substituting our equilibrium expressions into this equation:

$$ 0.84 = (0.50 + x) + x $$

$$ 0.84 = 0.50 + 2x $$

### Final Calculation

Now, it's just a matter of simple algebra to find $x$.

$$ 2x = 0.84 - 0.50 = 0.34 $$

$$ x = 0.17 	ext{ atm} $$

With $x$ known, we can determine the exact partial pressure of each gas at equilibrium:

$$ P_{	ext{NH}_3} = 0.50 + 0.17 = 0.67 	ext{ atm} $$

$$ P_{	ext{H}_2	ext{S}} = 0.17 	ext{ atm} $$

The equilibrium constant $K_p$ for this reaction only includes the gaseous products, because the active mass of a pure solid is constant and taken as unity.

$$ K_p = P_{	ext{NH}_3} 	imes P_{	ext{H}_2	ext{S}} $$

$$ K_p = 0.67 	imes 0.17 = 0.1139 $$

Rounding to two decimal places, we get **$0.11$**. This perfectly matches option (a). The beauty of this problem lies in carefully accounting for the initial conditions before applying the equilibrium constraints.

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Visualized Solution

\text{Initial State of the Flask}

\text{The Decomposition Reaction}

\text{Total Pressure at Equilibrium}

\text{Solving for } x

\text{Equilibrium Partial Pressures}

\text{Calculating } K_p

\text{Conclusion & Reflection}

The Sigma Insight: Law of Mass Action

Analyzing the Setup

The Master Equation

The Pressure Constraint

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Initial State of the Flask} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{The Decomposition Reaction} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Total Pressure at Equilibrium} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Solving for } x .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Equilibrium Partial Pressures} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating } K_p .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Conclusion & Reflection} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }