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) arrow 2 PBr 3(g) ; Δr H°=-243 kJ mol -1 will be

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) arrow 2 PBr 3(g) ; Δr H°=-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 .=30.97 g mol -1]

Tardigrade Admin · Accepted Answer

Molar mass of phosphorus =30.97 mol -1 2 P (s)+3 Br 2( l ) arrow 2 PBr 3(g), Δ H°=-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 =(243/30.97) kJ and 2.63 g of P when reacted liberates =(243/30.97) × 2.63=20.64 kJ Hence, the enthalpy change will be =20.64 kJ

Q. The enthalpy change when $2.63 g$ of phosphorus reacts with an excess of bromine according to the equation :
$2 P_{(s)} + 3 B r_{2 (l)} \to 2 PB r_{3 (g)}; Δ_{r} H^{\circ} = - 243 k J m o l^{- 1}$ will be

$103 k J$

$10.3 k J$

$20.6 k J$

$24.3 k J$ [Given: Molar mass of phosphorus $= 30.97 g m o l^{- 1}]$

Q. The enthalpy change when 2.63g of phosphorus reacts with an excess of bromine according to the equation : 2P(s)​+3Br2(l)​→2PBr3(g)​;Δr​H∘=−243kJmol−1 will be

103kJ

10.3kJ

20.6kJ

24.3kJ [Given: Molar mass of phosphorus =30.97gmol−1]

Q. The enthalpy change when $2.63 g$ of phosphorus reacts with an excess of bromine according to the equation :
$2 P_{(s)} + 3 B r_{2 (l)} \to 2 PB r_{3 (g)}; Δ_{r} H^{\circ} = - 243 k J m o l^{- 1}$ will be

$103 k J$

$10.3 k J$

$20.6 k J$

$24.3 k J$ [Given: Molar mass of phosphorus $= 30.97 g m o l^{- 1}]$