When a corpse decomposes, much of the phosphorus in the body is converted to pho

Question

When a corpse decomposes, much of the phosphorus in the body is converted to phosphine (PH3), a colourless gas with the odour of rotting fish. Phosphorus is a highly reactive molecule that ignites spontaneously in air. In the graveyard, phosphine that escapes from the ground ignites in air, giving small flashes of flame. These flashes are sometimes attributed to supernatural causes such as will-o -the-wisp. Determine delta h for the combustion of phosphine:

PH3(g)+O2(g) --> P4O10(s) + H20 (g).

i am not given any value for enthalpies of formation so how can i solve the question?

Explanation / Answer

Balanced equation for combustion reaction of phosphine is

4 PH3(g) + 8 O2(g)------> P4O10(s) + 6 H2O(g)

We know that

heat of formation of PH3 (g) = + 9.2 KJ/mol

heat of formation of O2 (g)  = 0 KJ/mol

heat of formation of P4O10(s) = -3012.5 KJ/mol

heat of formation of H2O(g) = -241.8 kJ/mol

Hence,

Enthalpy of combustion of phosphine ( H )

= Heat of formation of products - Heat of formation of reactants

= [ heat of formation of P4O10 + (6 x heat of formation of H2O)] - [ 4x (heat of formation of PH3) + 8x heat of formation of O2]

= [ ( -3012.5 KJ/mol) + (6 x -241.8 KJ/mol) ] - [ 4x (+9.2 KJ/mol) + 0 ]

= [ -3012.5 KJ/mol- 1450.8 KJ/mol ] - 36.8 KJ/mol

= -4500.1 KJ/mol

H = -4500.1 KJ/mol

Therefore, enthalpy of the combustion of phosphine, PH3(g) = -4500.1 KJ/mol

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