The standard heat of formation of PI3(s) is 24.7 kJ mol1 and the PI bond energy

Question

The standard heat of formation of PI3(s) is 24.7 kJ mol1 and the PI bond energy in this molecule is 184 kJ mol1. The standard heat of formation of P(g) is 334 kJ mol1 and that of I2(g) is 62 kJ mol1. The I2 bond energy is 151 kJ mol1. Calculate the heat of sublimation of PI3[PI3(s) PI3(g)] The standard heat of formation of PI3(s) is 24.7 kJ mol1 and the PI bond energy in this molecule is 184 kJ mol1. The standard heat of formation of P(g) is 334 kJ mol1 and that of I2(g) is 62 kJ mol1. The I2 bond energy is 151 kJ mol1. Calculate the heat of sublimation of PI3[PI3(s) PI3(g)]

Explanation / Answer

iverall reaction

P + 1.5 I2 ---> PI3

the process taking place in the reactions are ....

P (s) ----P (g) 334 kJ mol1

delta H I2 (g) = 62 kJ mol1

I2 ---2I-   151 kJ mol1

bond diss energy PI = 184 kJ mol1

delta Hf PI3 (g) = ? kJ mol1

delta Hf PI3 (s)= -24.7 kJ mol1

heat of sublimation of PI3[PI3(s) PI3(g)] = (delta Hf PI3 (g) - (delta Hf PI3 (s)

delta Hf PI3 (g) =

(P (s) ----P (g) ) + 1.5 (delta H I2 (g)) +( 0.5 I2 ---I-) + (3X bond diss energy PI )

= 334 + 1.5 X 62 + 1.5 X 151 + 3 X 184

= 1211.7 kJ mol1

hence

heat of sublimation of PI3[PI3(s) PI3(g)] = (delta Hf PI3 (s) - (delta Hf PI3) (g)

= -24.7 - 1211.7 kJ mol1

1236.4 KJ mol-1

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