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Dmitrij [34]
3 years ago
15

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

hanism for this overall reaction is shown below. No(g) + br2(g) br2(g) (fast step; keq = k1/k−1) k2 nobr(g) + no(g) → 2 nobr(g) (slow step) what is the rate law for formation of nobr in terms of reactants based on this mechanism
Chemistry
1 answer:
Svetach [21]3 years ago
4 0

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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q = -6464.9 kJ

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Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith
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The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

#SPJ1

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