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MrMuchimi
4 years ago
10

Experiments show that each of the following redox reactions is second-order overall: (1) NO2(g) + CO(g) → NO(g) + CO2(g) (2) NO(

g) + O3(g) → NO2(g) + O2(g) (a) When [NO2] in reaction 1 is doubled, the rate quadruples. Determine the rate law for this reaction. Rate = k[NO2][CO] k[NO2]2 k[NO2]3[CO]−1 k[NO2]4[CO]−2 (b) When [NO] in reaction 2 is doubled, the rate doubles. Determine the rate law for this reaction.
Chemistry
1 answer:
MatroZZZ [7]4 years ago
6 0

Answer:

a) rate law1 = k[NO2]²

b) rate law2 = k[NO][O3]

Explanation:

NO2(g) + CO(g) → NO(g) + CO2(g)

NO(g) + O3(g) → NO2(g) + O2(g)

When [NO2] in reaction 1 is doubled, the reaction quadruples

Rxn is second order.

rate law1= [NO2]^a [CO]^b

rate law1= [NO2]² [CO]^0

rate law1 = k[NO2]²

When [NO] in reaction 2 is doubled, the rate doubles.

Rxn is first order

The ratio is 1:1

this makes the rate law2 = k[NO][O3]

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7 0
3 years ago
What is the melting point of a substance?
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2 years ago
When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
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The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

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Have a nice day!

#LearnwithBrainly

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