Question

The enthalpy change when $2.63\, g$ of phosphorus reacts with an excess of bromine according to the equation :
$2 P _{(s)}+3 Br _{2(l)} \rightarrow 2 PBr _{3(g)} ; \Delta_{r} H^{\circ}=-243\, kJ\, mol ^{-1}$ will be

• A. $103\, kJ$
• B. $10.3\, kJ$
• C. $20.6\, kJ$
• D. $24.3\, kJ$ [Given: Molar mass of phosphorus $\left.=30.97 \,g\, mol ^{-1}\right]$

Answer 1

Answer: (C)

Molar mass of phosphorus $=30.97\, mol ^{-1}$

$2 P _{(s)}+3 Br _{2( l )} \rightarrow 2 PBr _{3(g)}, \Delta H^{\circ}=-243\, kJ mol ^{-1}$

We know, $30.97 g (1$ mole $)$ of $P$ when reacted liberates $243\, kJ$ of energy

$1 g$ of $P$ when reacted will liberate $=\frac{243}{30.97}\, kJ$

and $2.63 g$ of $P$ when reacted liberates

$=\frac{243}{30.97} \times 2.63=20.64\, kJ$

Hence, the enthalpy change will be $=20.64 \,kJ$