Red phosphorus reacts with liquid bromine in an exotermic reaction `:` `2P_((s))+3Br_(2(l))rarr2PBr_(3(g))` `Delta_(r)H^(@)=-243kJ`. Calculate the enthalpy change when `2.63g` of phosphorus with an excess of bromine in this way.

Explanation

Amount of phosphorous
`=m/M=(2.63 g)/(30.97 g mol^(-1))=0.0849 mol` of `P`
Here, enthalpy change of `-243 kJ` is associated with two moles of phosphorus, we can construct a conversion factor and use it to calculate the enthalpy change associated with `0.0849 mol` as
`DeltaH=0.0849 mol Pxx((-243 kJ)/(2 mol P))=-10.3 kJ`
At constant pressure, `10.3 kJ` of heat will be released for this amount `(0.0849 mol)` of phosphorus. The '2 mol P' appears in the denominator of the conversion factor because of the coefficient '2' with `P` in the balanced chemical equation.