Question

When red phosphorus burns in oxygen the enthalpy change is $- 2967 kJ {mol}^{- 1}$ . For white phosphorus the enthalpy change is $- 2984 kJ {mol}^{- 1}$ . For both forms of phosphorus the reaction taking place is: $P_{4} (s) + 5 O_{2} (g) \to P_{4} O_{10} (s)$ .

Use this information to calculate the enthalpy change for the transformation: $P_{4} (white) \to P_{4} (red)$ .

Accepted Answer

First of all the two forms of phosphorus are allotropes, which means when both are burnt they form the same compound. Since when white phosphorus gives off heat upon transformation into red phosphorus, the energy level of red phosphorus is lower than that of white phosphorus.Now remember the energy level of phosphorus oxide is lower than both, so we need to see which transition in energy level gives the most energy. Of course, white phosphorus falls from a higher energy level, so its combustion is more exothermic.

$\underset{White Phosphorus}{P_{4} (s)} + 5 O_{2} (g) \to P_{4} O_{10} (s); ∆ H_{f}^{°} = - 2984 kJ {mol}^{- 1} . ∆ H_{t}^{°} ↘ ↗ (- 2967 kJ {mol}^{- 1}) \underset{Red Phosphorus}{P_{4} (s)} + 5 O_{2} (g)$

$∆ H_{f}^{°} [P_{4} (W)] = ∆ H_{t}^{°} + ∆ H_{f}^{°} [P_{4} (R)] . - 2984 kJ {mol}^{- 1} = ∆ H_{t}^{°} - 2967 kJ {mol}^{- 1} . ∆ H_{t}^{°} = 2967 kJ {mol}^{- 1} - 2984 kJ {mol}^{- 1} . ∆ H_{t}^{°} = - 17 kJ {mol}^{- 1} .$

The enthalpy change for the transformation of white phosphorus into red phosphorus is $- 17 kJ {mol}^{- 1}$ .

Lawrie Ryan and Roger Norris Solutions for Chapter: Enthalpy Changes, Exercise 11: EXAM-STYLE QUESTIONS

Lawrie Ryan Chemistry Solutions for Exercise - Lawrie Ryan and Roger Norris Solutions for Chapter: Enthalpy Changes, Exercise 11: EXAM-STYLE QUESTIONS

Questions from Lawrie Ryan and Roger Norris Solutions for Chapter: Enthalpy Changes, Exercise 11: EXAM-STYLE QUESTIONS with Hints & Solutions

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