Boron can undergo the following reactions with the given enthalpy changes: $2…

Boron can undergo the following reactions with the given enthalpy changes:
$2 \mathrm{~B}(\mathrm{~s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s}) ; \Delta \mathrm{H}=-1260 \mathrm{~kJ}$
$2 \mathrm{~B}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g}) ; \Delta \mathrm{H}=30 \mathrm{~kJ}$
Assume no other reactions are occurring. If in a container (operating at constant pressure) which is isolated from the surrounding, mixture of $\mathrm{H}_{2}$ (gas) and $\mathrm{O}_{2}$ (gas) are passed over excess of $\mathrm{B}(\mathrm{s})$, then calculate the molar ratio $\left(\mathrm{O}_{2}: \mathrm{H}_{2}ight)$ so that temperature of the container do $\mathrm{n} \mathrm{o}$ change :
  1. $15: 3$
  2. $1: 42$
  3. $42: 1$
  4. $1: 84$

Solution

No. of moles of $\mathrm{O}_{2}$ required to supplied 30 $\mathrm{kJ}$ heat to second reaction
$=\frac{30}{1260} \times \frac{3}{2}=\frac{1}{28}$
$\mathrm{n}_{\mathrm{O}_{2}}: \mathrm{n}_{\mathrm{H}_{2}}=\frac{1}{28}: 3$ or $1: 84$ ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more THERMODYNAMICS questions on Aicharya