Question #51b13

1 Answer
May 18, 2017

Answer:

#W=-101325*1.248*10^-3=-126.5" J"#

#"∆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(NaN3)=2.25/65.02=0.0346" mol"#
#n(N2)=3/2*"n(NaN3)"=0.0346*3/2=0.0519" mol"#
#m(N2)=n(N2)*M(N2)=0.0519*28.02=1.454" g"#
#V(N2)=m/"density"=1.454/1.165=1.248" L"#
#V(N2)=1.248*10^-3" 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*1.248*10^-3=-126.5" Pa m"^3=-126.5" J"#

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

#1" J" = 1" Pa m"^3#.

The heat released is the enthalpy change:

#"∆H"=-2340" J"#

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

#"∆H"="∆E"+"P∆V"#
#"∆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.