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Vsevolod [243]
1 year ago
7

The decomposition of nitramide, O 2 NNH 2 , O2NNH2, in water has the chemical equation and rate law O 2 NNH 2 ( aq ) ⟶ N 2 O ( g

) + H 2 O ( l ) rate = k [ O 2 NNH 2 ] [ H + ] O2NNH2(aq)⟶N2O(g)+H2O(l)rate=k[O2NNH2][H+] A proposed mechanism for this reaction is O 2 NNH 2 ( aq ) k 1 ⇌ k − 1 O 2 NNH − ( aq ) + H + ( aq ) ( fast equilibrium ) O2NNH2(aq)⇌k−1k1O2NNH−(aq)+H+(aq)(fast equilibrium) O 2 NNH − ( aq ) k 2 −→ N 2 O ( g ) + OH − ( aq ) ( slow ) O2NNH−(aq)→k2N2O(g)+OH−(aq)(slow) H + ( aq ) + OH − ( aq ) k 3 −→ H 2 O ( l ) ( fast ) H+(aq)+OH−(aq)→k3H2O(l)(fast) What is the relationship between the observed value of k k and the rate constants for the individual steps of the mechaanism?
Chemistry
1 answer:
valkas [14]1 year ago
5 0

Answer:

Explanation:

The given overall reaction is as follows:

O 2 N N H ₂( a q ) k → N ₂O ( g ) + H ₂ O( l )

The reaction mechanism for this reaction is as follows:

O ₂ N N H ₂ ⇌ k 1 k − 1  O ₂N N H ⁻ + H ⁺ ( f a s t  e q u i l i b r i u m )

O ₂ N N H − k ₂→ N ₂ O + O H ⁻ ( s l ow )

H ⁺ + O H − k ₃→ H ₂ O ( f a s t )

The rate law of the reaction is given as follows:

k = [ O ₂ N N H ₂ ]  / [ H ⁺ ]

The rate law can be determined by the slow step of the mechanism.

r a t e = k ₂ [ O ₂ N N H ⁻ ] . . . ( 1 )

Since, from the equilibrium reaction

k e q = [ O ₂ N N H ⁻ ] [ H ⁺ ] /[ O ₂ N N H ₂ ] = k ₁ /k − 1

[ O ₂ N N H ⁻] = k ₁ /k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]. . . . ( 2 )

Substitituting the value of equation (2) in equation (1) we get.

r a t e = k ₂ k ₁/ k − 1  × [ O ₂ N N H ₂ ] /[ H ⁺ ]

Therefore, the overall rate constant is

k = k₂k₁/k-1

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A bottle of antiseptic hydrogen peroxide (H2O2) is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solutio
PolarNik [594]

Answer:

= 12 mL H202

Explanation:

Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v

It implies that, 3ml  H2O2 is present in 100 ml of solution.

Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;

we have;

 (3.0 mL/ 100 mL) × 400 mL

= 12 mL H202

6 0
2 years ago
Ethyl alcohol is produced by the fermentation of glucose, C6H12O6. C6H12O6 (s) → 2 C2H5OH (l) + 2 CO2 (g) ΔH° = – 69.1 kJ Given
Norma-Jean [14]

Answer:

-1273.3

Explanation:

Enthalpy of formation of a compound is the amount of heat absorbed or evolved when one mole of the compound is formed from other compounds.

enthalpy of formation Of CO2 = 2 X -393.5 = -787

enthalpy of formation Of C2H5OH = 2 X -277.7 = -555.4

enthalpy of formation Of C6H12O6 = 69.1 (reverse sign) + (-787 + -555.4) = - 1273.3 Joules

6 0
2 years ago
a student adds 3.5 moles of solute to enough water to make a 1500mL solution. what is the concentration?
aksik [14]
<h2>Hello!</h2>

The answer is:

MolarConcentration=\frac{3.5moles}{volume(1.5L)}=2.33molar

<h2>Why?</h2>

Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:

MolarConcentration=\frac{mass(solute)}{volume(solution)}

Now, we know that the mass of the solute is equal  3.5 moles and the volume is equal to 1500 mL or 1.5L

Then, substituting into the equation, we have:

MolarConcentration=\frac{3.5moles}{1.5L}=2.33molar

Have a nice day!

7 0
1 year ago
Read 2 more answers
Plseas help Calculate the amount of moles in 57.6 Liters of Carbon Dioxide.
slavikrds [6]
Convert 57.6 L to dm3 and divide it by 24
8 0
1 year ago
What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Anna11 [10]
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

hope this helps!
5 0
2 years ago
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