The values of work, w, for the system if 16.5 g NaN 3 reacts completely at 1.00 atm and 22 ∘ C. The work done is -919 J.
The equation of the reaction is;
2 NaN 3 ( s ) ⟶ 2 Na ( s ) + 3 N 2 ( g )
Number of moles of NaN3 = 16.5 g/65 g/mol = 0.25 moles
If 2 moles of NaNa3 yields 3 moles of N2
0.25 moles of NaN3 yields 0.25 moles × 3 moles/2 moles
= 0.375 moles of N2
We need to find the volume change using the;
PV = nRT
P = 1.00 atm
V =?
n = 0.375 moles of N2
R = 0.082 atmLK-1mol-1
T = 22 ∘ C + 273 = 295 K
V = nRT/P
V = 0.375 × 0.082 × 295/ 1.00
V = 9.07 L
Recall that during expansion the gas does work. Work done by the gas is;
W = -PΔV
W =-( 1 atm × 9.07 L)
W = -9 Atm
Again;
1 L atm = 101.325 J
So,
-9 atmL = -9 atmL × 101.325 J/1 L atm
= -919 J
The work done is -919 J.
Learn more about work done here:-brainly.com/question/25573309
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Automobile airbags contain solid sodium azide, NaN 3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2 NaN 3 ( s ) ⟶ 2 Na ( s ) + 3 N 2 ( g ) Calculate the value of work, w, for the system if 16.5 g NaN 3 reacts completely at 1.00 atm and 22 ∘ C.
