24. A laboratory chemist wants to produce 25.0 g NO2 by the following reactions. If the
1 answer:
The amount of N2O5 to start with would be 35.79 grams
<h3>Stoichiometric calculations</h3>
From the balanced equation of the reactions:
Mole ratio of N2O5 and NO2 = 1:2
Since the reaction's actual yield is 82%, the theoretical yield would be: 25 x 100/82 = 30.49 grams
Mole of 25 g NO2 = 30.49/46
= 0.66 mole
Equivalent mole of N2O5 = 0.54/2 mole
= 0.33 moles
Mass of 0.27 mole N2O5 = 0.33 x 108.01
= 35.79 grams
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Answer:
B. First order, Order with respect to C = 1
Explanation:
The given kinetic data is as follows:
A + B + C → Products
[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)
1. 0.4 0.4 0.2 160
2. 0.2 0.4 0.4 80
3. 0.6 0.1 0.2 15
4. 0.2 0.1 0.2 5
5. 0.2 0.2 0.4 20
The rate of the above reaction is given as:
where x, y and z are the order with respect to A, B and C respectively.
k = rate constant
[A], [B], [C] are the concentrations
In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.
Order w.r.t A : Use trials 3 and 4
Order w.r.t B : Use trials 2 and 5
Order w.r.t C : Use trials 1 and 2
we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:
z = 1
Therefore, order w.r.t C = 1