# Question 51b13

May 18, 2017

$W = - 101325 \cdot 1.248 \cdot {10}^{-} 3 = - 126.5 \text{ J}$

$\text{∆E"="∆H"-"P∆V"=-2340-126.5=-2466.5" J}$

#### Explanation:

First of all, always use SI units to get the answer out in SI units. In this case, we want work and ∆E in joules, so use volume in m^3 and pressure in Pa.
You need the volume that the N2 gas occupies:

$n \left(N a N 3\right) = \frac{2.25}{65.02} = 0.0346 \text{ mol}$
$n \left(N 2\right) = \frac{3}{2} \cdot \text{n(NaN3)"=0.0346*3/2=0.0519" mol}$
$m \left(N 2\right) = n \left(N 2\right) \cdot M \left(N 2\right) = 0.0519 \cdot 28.02 = 1.454 \text{ g}$
$V \left(N 2\right) = \frac{m}{\text{density"=1.454/1.165=1.248" L}}$
$V \left(N 2\right) = 1.248 \cdot {10}^{-} 3 {\text{ m}}^{3}$

Note that the Ideal Gas Law will give you the same volume.

As the initial volume is zero:

∆V="Vf"-"Vi"=1.248*10^-3-0=1.248*10^-3" m"^3

Now substitute the external pressure (convert 1 atm to Pa) and the change in volume into the expression for work done:

W = -P∆V#
$W = - 101325 \cdot 1.248 \cdot {10}^{-} 3 = - 126.5 \text{ Pa m"^3=-126.5" J}$

By using SI units, the answer has come out neatly in J, as

$1 {\text{ J" = 1" Pa m}}^{3}$.

The heat released is the enthalpy change:

$\text{∆H"=-2340" J}$

Rearranging the formula for enthalpy allows us to solve for the internal energy change:

$\text{∆H"="∆E"+"P∆V}$
$\text{∆E"="∆H"-"P∆V"=-2340-126.5=-2466.5" J}$

From the system's point of view, this means that during the reaction 2340 J was transferred as heat to the surroundings and a further 126.5 J was expended as expansion work, giving a total internal energy change/loss (∆E) of 2466.5 J.