Question

How much energy (in joules) will be given off from the following reaction, if 3.0g of sodium reacts with 1.2g of nitrogen in the following reaction? 2Na+ N2 ——> 2NaN3 delta H= -42.7kJ

Answer 1

In the reaction

2Na + 3N₂ $\rightarrow$ 2NaN₃,

2 moles of sodium react with 3 moles of nitrogen:

$n(Na)/2 = n(N_2)/3$ .

The number of the moles contained in 3 g of sodium is its mass divided by its molar mass $M$ (22.99 g/mol):

$n(Na) = \frac{m(Na)}{M(Na)} = \frac{3.0}{22.99} = 0.1305$ mol.

The number of the moles of nitrogen can also be calculated from its mass ($M_{N_2} =$ 28.01 g/mol):

$n(N_2) = \frac{m(N_2)}{M(N_2)} = \frac{1.2}{28.01} = 0.04284$ mol.

Let's compare the equivalent number of the moles of sodium and of nitrogen :

$3·n(Na)= 0.3915; 2·n(N_2) = 0.08568 \Rightarrow 3n(Na)>2n(N_2)$ .

Therefore, there is an excess of sodium. It means that in order to calculate the energy released in the system, we must use the number of the moles of nitrogen. As 3 moles of nitrogen are used in the reaction, 3 moles of nitrogen are equivalent to the release of -42.7 kJ. So, the energy released is the product of the molar enthalpy and the number of the moles of nitrogen, divided by the conversion factor for nitrogen ( 3 mol N₂) :

$Q = ∆H·n(N_2)/3 = 42.7·0.04284/3 = -0.610$ kJ.

Answer: -0.610 kJ.