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1 year ago

Automobile airbags contain solid sodium azide, nan3, that reacts to produce nitrogen gas when heated, thus inflating the bag.

1 answer:

5 0

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.

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For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

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Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

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$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

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