Answer
a. 4.51 kJ are required to produce 3.00 g of NO
b. This is the complete equation for the decomposition of NO:
$$2 NO(g) \longrightarrow N_2(g)+O_2(g)+90.2 \space kJ$$
c. 7.51 kJ are released when 5.00 g of NO decomposes.
Work Step by Step
a.
- Find the conversion factors:
$ NO $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 30.01 g/mol
$$ \frac{1 \space mole \space NO }{ 30.01 \space g \space NO } \space and \space \frac{ 30.01 \space g \space NO }{1 \space mole \space NO }$$
According to the balanced equation:
$$ \frac{ 2 \space moles \space NO }{ 90.2 \space kJ} \space and \space \frac{ 90.2 \space kJ}{ 2 \space moles \space NO }$$
- Find the total energy using these conversion factors:
$$ 3.00 \space g \space NO \times \frac{1 \space mole \space NO }{ 30.01 \space g \space NO } \times \frac{ 90.2 \space kJ}{ 2 \space moles \space NO } = 4.51 \space kJ$$
b. Since $\Delta H \gt 0$, the reaction is endothermic, thus, the heat will appear in the reactants side:
$$N_2(g)+O_2(g)+90.2 \space kJ \longrightarrow 2 NO(g)$$
But, this is the reaction for the formation of NO. If we want to write the reaction for the decomposition of NO, we can just reverse the equation:
$$2 NO(g) \longrightarrow N_2(g)+O_2(g)+90.2 \space kJ$$
c.
- Find the conversion factors:
$ NO $ : ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 30.01 g/mol
$$ \frac{1 \space mole \space NO }{ 30.01 \space g \space NO } \space and \space \frac{ 30.01 \space g \space NO }{1 \space mole \space NO }$$
According to the balanced equation:
$$ \frac{ 2 \space moles \space NO }{ -90.2 \space kJ} \space and \space \frac{ -90.2 \space kJ}{ 2 \space moles \space NO }$$
- Find the total energy using these conversion factors:
$$ 5.00 \space g \space NO \times \frac{1 \space mole \space NO }{ 30.01 \space g \space NO } \times \frac{ -90.2 \space kJ}{ 2 \space moles \space NO } = -7.51 \space kJ$$
** Since the heat is now in the products side, it is negative.
** Negative heat = released heat
