When a certain amount of ethylene was burnt 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ , the volume of O 2 (at NTP) that entered into the reaction is

Question

When a certain amount of ethylene was burnt 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ , the volume of O 2 (at NTP) that entered into the reaction is (A) 296.5 mL (B) 296.5 L (C) 6226 × 22.4 mL (D) 22.4 L

Tardigrade Admin · Accepted Answer

Combustion of ethylene C 2 H 4+3 O 2 arrow 2 CO 2+2 H 2 O ; Δ H=-1411 kJ To evolve 1411 kJ heat, 3 moles i.e., 3 × 22.4 L oxygen is used. Volume of oxygen used to evolve 6226 kJ heat =(3 × 22.4/1411) × 6226=296.5 L

Q. When a certain amount of ethylene was burnt $6226 k J$ heat was evolved. If heat of combustion of ethylene is $1411 k J,$ the volume of $O_{2}$ (at NTP) that entered into the reaction is

$296.5 m L$

$296.5 L$

$6226 \times 22.4 m L$

$22.4 L$

Combustion of ethylene

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O; Δ H = - 1411 k J$

To evolve $1411 k J$ heat, $3$ moles i.e.,

$3 \times 22.4 L$ oxygen is used. Volume of oxygen used to evolve $6226 k J$ heat

$= \frac{3 \times 22.4}{1411} \times 6226 = 296.5 L$

Q. When a certain amount of ethylene was burnt 6226kJ heat was evolved. If heat of combustion of ethylene is 1411kJ, the volume of O2​ (at NTP) that entered into the reaction is

296.5mL

296.5L

6226×22.4mL

22.4L