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topjm [15]
3 years ago
7

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is tho

ught to be:
NO2 + NO2 -------> NO3 + NO
NO3 + CO ---------> NO2 + CO2
1. Write the rate law expected for this mechanism. What is the overall balanced equation of the reaction?
Chemistry
1 answer:
Dahasolnce [82]3 years ago
3 0

Explanation:

NO_2 + NO_2 \rightarrow NO_3 + NO (slow)..(1)

NO_3 + CO\rightarrow NO_2 + CO_2 (fast)..(2)

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

The reaction occurring in more than 1 step , the rate of the reaction is determined from the reaction which is occurring at slow speed.

So, the rate of reaction for the given mechanism will be determine by step 1.

Rate law expression of the given mechanism is :

R=k[NO_2][NO_2]

The rate law expected for this mechanism:

R=K[NO_2]^2

The overall balanced equation of the reaction :

=  [1] + [2]

NO_2+CO\rightarrow CO_2+NO

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2 years ago
A brine solution of salt flows at a constant rate of 9 ​L/min into a large tank that initially held 100 L of brine solution in w
Valentin [98]

Answer:

a) C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) The concentration of salt in the tank attains the value of 0.01 kg/L at time, t = 0.0713 min = 4.28s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the Concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F

Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dC/dt = FᵢCᵢ - FC

Fᵢ = 9 L/min, Cᵢ = 0.02 kg/L, F = 9 L/min

dC/dt = 0.18 - 9C

dC/(0.18 - 9C) = dt

∫ dC/(0.18 - 9C) = ∫ dt

(-1/9) In (0.18 - 9C) = t + k

In (0.18 - 9C) = -9t - 9k

-9k = K

In (0.18 - 9C) = K - 9t

At t = 0, C = 0.1/100 = 0.001 kg/L

In (0.18 - 9(0.001)) = K

In 0.171 = K

K = - 1.766

So, the equation describing concentration of salt at anytime in the tank is

In (0.18 - 9C) = -1.766 - 9t

In (0.18 - 9C) = - (9t + 1.766)

0.18 - 9C = e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾

9C = 0.18 - (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) when C = 0.01 kg/L

0.01 = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

0.09 = (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

- (9t + 1.766) = In 0.09

- (9t + 1.766) = -2.408

(9t + 1.766) = 2.408

9t = 2.408 - 1.766 = 0.642

t = 0.642/9 = 0.0713 min = 4.28s

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Explanation:

Learning Task 3. Solving word problems involving multiplication of deci-

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regular rate for overtime and double the rate for a holiday. How much

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How much does he save in a month?

4. Mr. Fernandez has a monthly pay of P 5,450. The tax deducted from

year?

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