Car airbags inflate upon sudden deceleration due to the following chemical reaction:
a) What is the value of ??? ?? 289 for NaN 3 ?
b) A fully inflated airbag requires 2.4 mol of N 2 (g). Calculate the enthalpy change to fully inflate an airbag, which occurs when an airbag is fully inflated.
c) Assuming all energy goes into heating the 2.4 mol of N 2 (g), what will the final temperature of the gas be? C p for N 2 (g) = 29.1 J K –1 mol -1
Solution
Solution :-
NaN3(s) ----> Na(s) + 3/2 N2(g)Â Â Â Â delta H = -21.7 kJ/mol
#a)Calculating the   value of delta Hfo of the NaN3(s)
Delta Hrxn = sum of delta Hfo product - sum of delta Hfo reactant
-21.7 kJ/mol= [(1*Na)+(3/2*N2)]-[1*NaN3]
-21.7 kJ/mol = [(1*0)+(3/2*0)] - [1*NaN3]
-21.7 kJ/mol = 0 - [1*NaN3]
21.7 kJ/mol = NaN3(s)
Therefore the Delta Hfo of the NaN3(s) is 21.7 kJ/mol
#b) fully inflate airbag contains 2.4 mol N2(g)
As we know from the reaction equation (3/2) mol N2 gives -21.7 kJ energy
So lets calculate the amount of energy given for the 2.4 mol N2
2.4 mol N2 * -21.7 kJ / (3/2) mol N2 = -34.72 kJ
Therefore enthalpy change for the fully inflate airbag is -34.72 kJ
#c) initial temperature of the gas should be 298 K
Lets calculate the change in the temperature after
Delta T = delta H/(n* Cp)
Delta T= (34.72 kJ*1000 J/1kJ)/(2.4 mol * 29.1 J/mol K)
= 497 K
So the final temperature of the N2 gas will be 497 K
.
Car airbags inflate upon sudden deceleration due to the following chem.docx
1. Car airbags inflate upon sudden deceleration due to the following chemical reaction:
a) What is the value of ??? ??
289 for NaN 3 ?
b) A fully inflated airbag requires 2.4 mol of N 2 (g). Calculate the enthalpy change to fully
inflate an airbag, which occurs when an airbag is fully inflated.
c) Assuming all energy goes into heating the 2.4 mol of N 2 (g), what will the final temperature
of the gas be? C p for N 2 (g) = 29.1 J K –1
mol -1
Solution
Solution :-
NaN3(s) ----> Na(s) + 3/2 N2(g)Â Â Â Â delta H = -21.7 kJ/mol
#a)Calculating the   value of delta Hfo of the NaN3(s)
Delta Hrxn = sum of delta Hfo product - sum of delta Hfo reactant
-21.7 kJ/mol= [(1*Na)+(3/2*N2)]-[1*NaN3]
-21.7 kJ/mol = [(1*0)+(3/2*0)] - [1*NaN3]
-21.7 kJ/mol = 0 - [1*NaN3]
21.7 kJ/mol = NaN3(s)
Therefore the Delta Hfo of the NaN3(s) is 21.7 kJ/mol
#b) fully inflate airbag contains 2.4 mol N2(g)
As we know from the reaction equation (3/2) mol N2 gives -21.7 kJ energy
So lets calculate the amount of energy given for the 2.4 mol N2
2. 2.4 mol N2 * -21.7 kJ / (3/2) mol N2 = -34.72 kJ
Therefore enthalpy change for the fully inflate airbag is -34.72 kJ
#c) initial temperature of the gas should be 298 K
Lets calculate the change in the temperature after
Delta T = delta H/(n* Cp)
Delta T= (34.72 kJ*1000 J/1kJ)/(2.4 mol * 29.1 J/mol K)
= 497 K
So the final temperature of the N2 gas will be 497 K
![Car airbags inflate upon sudden deceleration due to the following chemical reaction:
a) What is the value of ??? ??
289 for NaN 3 ?
b) A fully inflated airbag requires 2.4 mol of N 2 (g). Calculate the enthalpy change to fully
inflate an airbag, which occurs when an airbag is fully inflated.
c) Assuming all energy goes into heating the 2.4 mol of N 2 (g), what will the final temperature
of the gas be? C p for N 2 (g) = 29.1 J K –1
mol -1
Solution
Solution :-
NaN3(s) ----> Na(s) + 3/2 N2(g)Â Â Â Â delta H = -21.7 kJ/mol
#a)Calculating the   value of delta Hfo of the NaN3(s)
Delta Hrxn = sum of delta Hfo product - sum of delta Hfo reactant
-21.7 kJ/mol= [(1*Na)+(3/2*N2)]-[1*NaN3]
-21.7 kJ/mol = [(1*0)+(3/2*0)] - [1*NaN3]
-21.7 kJ/mol = 0 - [1*NaN3]
21.7 kJ/mol = NaN3(s)
Therefore the Delta Hfo of the NaN3(s) is 21.7 kJ/mol
#b) fully inflate airbag contains 2.4 mol N2(g)
As we know from the reaction equation (3/2) mol N2 gives -21.7 kJ energy
So lets calculate the amount of energy given for the 2.4 mol N2](https://image.slidesharecdn.com/carairbagsinflateuponsuddendecelerationduetothefollowingchem-230130170151-81f76460/85/car-airbags-inflate-upon-sudden-deceleration-due-to-the-following-chemdocx-1-320.jpg)
