Answer:
0.0253 M/s
Explanation:
From the reaction
N₂ + 3H₂ → 2NH₃
The rate of reaction can be written as
Rate = - ![\frac{d[N_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BN_2%5D%7D%7Bdt%7D)
= - ![\frac{1}{3} \frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
= + ![\frac{1}{2} \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
From the above rate equation we can conclude that the rate of reaction of N₂ is equal to one third of the rate of reaction of H₂,
So,
Rate of reaction of molecular nitrogen = 
Upon calculation, we get rate of reaction of molecular nitrogen = 0.0253 M/s
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Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
