Answer:
0.41kg/sec
Explanation:
PV= nRT
Given : V= 505 L
P=0.88 atm
R= 0.08206 Latm/K*mol
T= 172 .0C = 172+273 = 445 K
n = PV /RT = 0.88 * 505 / 0.08206 * 445 = 12.17 moles per sec of N2 are consumed
As per reaction : N2 + 3H2 ----> 2NH3
1 mole N2 is consumed to produce 2 moles NH3
moles of NH3 produced per sec :
(2 moles NH3/1mol N2) * 12.17 moles N2 = 24.34 moles NH3 per sec
grams of NH3 produced per sec =
24.34 moles NH3 per sec * molar mass NH3 = 24.34 moles NH3 per sec * 17.031 g/mol = 414.5 g NH3 per sec
rate in Kg/sec = 414.5 g NH3 per sec * (1kg /1000g) = 0.4145 Kg/sec
= 0.41kg/sec
Answer:
457.5kPa
Explanation:
Given data
V1=V2=350mL (<em>fixed volume</em> )
P1=366kPa
T1= 88 degrees Celsius
P2=??
T2= 110 degrees Celsius
For the general gas equation
P1V1/T1= P2V2/T2
V1=V2
P1/T1= P2/T2
Substitute
366/88= P2/110
Cross multiply we have
P2*88=366*110
P2*88= 40260
P2= 40260/88
P2= 457.5 kPa
Hence the pressure will change to 457.5kPa
Answer:
-2024 kJ
Explanation:
The combustion of propane is
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
And, ΔG° = ΔH° - TΔS°
Where T is the temperature in K.
ΔH° = ∑n*H°f, reactants - ∑n*H°f, products
Where n is the number of moles in the stoichiometry reaction. H°f, O₂(g) = 0 because it's a substance formed by only one element.
ΔH° = [4*(-226) +3*(-398)] - [-100] = -1998 kJ
ΔS° = ∑n*S°, reactants - ∑n*S°, products
ΔS° = [4*(184) + 3*214] - [5*204 + 271] = 87 J/K = 0.087 kJ/K
So
ΔG° = -1998 - 298*0.087
ΔG° = -2024 kJ
