Chemistry, 05.05.2020 18:32, oliviapolerop7209

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[O2NN H2][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)(f ast 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?

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answered: helpmewithmath70

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

answered: Guest

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answered: Guest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o