Question

learn by doing
refer to the reactions below to answer the following questions:
reaction 1 : ( 2 mathrm { b } ( s ) + \frac { 3 } { 2 } mathrm { o } _ { 2 } ( g )
ightarrow mathrm { b } _ { 2 } mathrm { o } _ { 3 } ( s ) ) ( delta h = - 1273 mathrm { kj }
reaction 2 : ( mathrm { ch } _ { 4 } ( g ) + 2 mathrm { o } _ { 2 } ( g )
ightarrow mathrm { co } _ { 2 } ( g ) + 2 mathrm { h } _ { 2 } mathrm { o } ( l ) ) ( delta h = - 890 mathrm { kj }
reaction 3 : ( mathrm { n } _ { 2 } ( g ) + 2 mathrm { o } _ { 2 } ( g )
ightarrow 2 mathrm { no } _ { 2 } ( g ) ) ( delta h = 68 mathrm { kj }
reaction 4 : ( mathrm { h } _ { 2 } ( g ) + \frac { 1 } { 2 } mathrm { o } _ { 2 } ( g )
ightarrow mathrm { h } _ { 2 } mathrm { o } ( g ) ) ( delta h = - 242 mathrm { kj }
which of the reactions is endothermic?
in which reaction would ( delta h = - 445 mathrm { kj } ) for the reaction of 1 mole of ( mathrm { o } _ { 2 } ) ?
which of the reactions would yield the same value for ( delta h ) if ( mathrm { o } _ { 2 } ( l ) ) reacted instead of ( mathrm { o } _ { 2 } ( g ) ) ?

Explanation:

Step1: Determine endothermic reaction

Endothermic reactions have $\Delta H>0$. Among the given reactions, only reaction 3 has $\Delta H = 68\ kJ>0$.

Step2: Find reaction with $\Delta H=-445\ kJ$ for 1 mole of $O_2$

For reaction 2: $\Delta H=-890\ kJ$ for 2 moles of $O_2$. So for 1 mole of $O_2$, $\Delta H=\frac{-890}{2}=-445\ kJ$.

Step3: Check reaction unaffected by $O_2$ state

In reaction 1, $B$ is solid and $B_2O_3$ is solid. The state of $O_2$ (gas or liquid) does not affect the $\Delta H$ as there is no phase - change of $O_2$ within the reaction context (unlike reactions where $O_2$ is a reactant in a way that its phase matters for the overall energy change).

Answer:

1. Reaction 3 is endothermic.
2. Reaction 2 has $\Delta H=-445\ kJ$ for the reaction of 1 mole of $O_2$.
3. Reaction 1 would yield the same value for $\Delta H$ if $O_2(l)$ reacted instead of $O_2(g)$.