Boron can undergo the following reactions with the given enthalpy changes :
$$\eqalign{
& 2B\left( s \right) + \frac{3}{2}{O_2}\left( g \right) \to {B_2}{O_3}\left( s \right);\,\Delta H = - 1260\,kJ \cr
& 2B\left( s \right) + 3{H_2}\left( g \right) \to {B_2}{H_6}\left( g \right);\,\Delta H = 30\,kJ \cr} $$
Assume no other reactions are occurring.
If in a container ( operating at constant pressure ) which is isolated from the surrounding, mixture of $${H_2}$$ (gas) and $${O_2}$$ (gas) are passed over excess of $$B(s),$$ then calculate the molar ratio $$\left( {{O_2}:{H_2}} \right)$$ so that temperature of the container do not change :
A.
15 : 3
B.
42 : 1
C.
1 : 42
D.
1 : 84
Answer :
1 : 84
Solution :
No. of moles of $${O_2}$$ required to supplied $$30\,kJ$$ heat to second reaction
$$\eqalign{
& = \frac{{30}}{{1260}} \times \frac{3}{2} = \frac{1}{{28}} \cr
& {n_{{O_2}}}:{n_{{H_2}}} = \frac{1}{{28}}:3\,\,\,{\text{or}}\,\,1:84 \cr} $$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$