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Red phosphorus reacts with liquid bromin...

Red phosphorus reacts with liquid bromine in an exothermic reaction
`2P(s) + 3 Br_(2) (l) rarr PBr_(3) (g), Delta_(r) H^(@) = - 243 kJ mol^(-1)`
Calculated the enthalpy change when `2.63 g` of phosphorus reacts with an excess of bromine in this way.

A

`10.3 kJ`

B

`1536 kJ`

C

`7.5 kJ`

D

`20.3 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
Calculate the number of moles of `P` and mulitply it with `Delta_(r) H^(@)` for 1 mol of `P`.
`n_(P) = (Mass_(P))/(Molar mass_(P)) = (2.63 g)/(30.97 mol^(-1)) = 0.0849 of P`
`Delta H = (0.0849 mol) ((-243 kJ)/(2 mol P))`
`= 10.3 kJ`
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