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Phosphorus tribromide decomposes to form phosphorus and bromine, as shown in the equation:

\[4 \text{PBr}_3(g) \rightarrow \text{P}_4(g) + 6 \text{Br}_2(g)\]

A chemist finds that at a certain temperature, the equilibrium mixture of phosphorus tribromide, phosphorus, and bromine has the following composition:

| Compound | Pressure at Equilibrium (atm) |
|----------|-------------------------------|
| \(\text{PBr}_3\) | 97.4 |
| \(\text{P}_4\) | 99.2 |
| \(\text{Br}_2\) | 97.2 |

Calculate the value of the equilibrium constant \(K_p\) for this reaction. Round your answer to two significant digits.

Answer :

Final answer:

The equilibrium constant (Kp) for the decomposition of phosphorus tribromide (PBr3) can be calculated using the equilibrium pressures and the equation for Kp, resulting in a value of 1.1 × 10⁵ (rounded to two significant digits).

Explanation:

The equilibrium constant, Kp, for the decomposition of phosphorus tribromide (PBr3) can be calculated using the equation:

Kp = (PP4) × (PBr2)⁶ / (PPBr3)⁴

Where 'P' denotes the partial pressure of each substance at equilibrium. Given the pressures at equilibrium for PBr3 (97.4 atm), P4 (99.2 atm), and Br2 (97.2 atm), the equation becomes:

Kp = (99.2 atm) × (97.2 atm)⁶ / (97.4 atm)⁴

Upon performing the calculations, we get:

Kp = 1.1 × 10⁵ (rounded to two significant digits)

Therefore, the equilibrium constant for the reaction at the given conditions is 1.1 × 10⁵.

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Rewritten by : Barada

Answer:

Kp = 9.3x10⁴

Explanation:

The reaction is this:

4PBr₃ (g) ⇄ P₄ (g) + 6Br₂ (g)

We define Kp from the partial pressures in equilibrium

PBr₃ → 97.4 atm

P₄ → 99.2 atm

Br₂ → 97.2 atm

Kp = (Partial pressure Br₂)⁶ . (Partial pressure P₄) / (Partial pressure PBr₃)⁴

Kp = 97.2⁶ . 99.2 / 97.4⁴

Kp = 929551.4533

929551.4533 → 9.3x10⁴