Question

1. What would a valid expression for the rate of the reaction below? and why?

4 NH3 + 7 O2 → 4 NO2 + 6 H2O

2. What is the mole fraction of a 4.0 M NaOH solution? The density of the solution is 1.04g/mol? (The answer is either 0.076 or 0.92 since the question didnt specify mole fraction of solute or solvent)

Answer 1

1.

Since no other data is added

we typically take a "elemtnary rate law"

which is based on cncentrations of reactants, and their coefficients

that is:

Rate = k*[NH3]^4[O2]^7

2.

mole frac = mol NaOH / total mol

assume a basis of 1 L solution

then

mass = V*D = 1*1040 = 1040 g

mol of NaOH = 4 mol

mas s= mol*MW = 4*40 = 160 g

then

mass of water = 1040-160 = 880

mol of water = mass/WM = 880/18 = 48.88888

total mol = 48.88888 + 44 = 52.88

mol frac of NaOH = 4/52.88 = 0.07564

for water = 1-x = 1-0.07564 = 0.92436