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. $6226 \times 22.4\, mL$
• D. $22.4\, L$

Answer 1

Answer: (B)

Combustion of ethylene

$C _{2} H _{4}+3 O _{2} \rightarrow 2 CO _{2}+2 H _{2} O ; \Delta H=-1411 \, kJ$

To evolve $1411 \,kJ$ heat, $3$ moles i.e.,

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

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