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Goshia [24]
3 years ago
9

What is an elementary reaction? 3.5. Given the reaction 2NO, + 1/202 = N,Os, what is the relationship between the rates of forma

tion of 1,0, and disappearance of the two reactants?
Chemistry
1 answer:
Alik [6]3 years ago
7 0

Answer:

a) Elementary reactions are those which proceed in single step only.

b) -\dfrac{1}{2}\dfrac{d\left [ NO_{2} \right ]}{dt}\ =\ -2\dfrac{d\left [ O_{2} \right ]}{dt}\ =\ \dfrac{1}{2}\dfrac{d\left [ N_{2}O_{5} \right ]}{dt}

Explanation:

a) In case of elementary reactions, there is only one transition sate formed during the reactions.

There are no intermediates formed in elementary reactions as they occur in one step.

b) For the reaction

2NO_{2}\ +\ \dfrac{1}{2}O_{2}\ \rightarrow N_{2}O_{5}

Relationship between rate of formation of products and rate of disappearance of reactants is given by rate law expression.

-\dfrac{1}{2}\dfrac{d\left [ NO_{2} \right ]}{dt}\ =\ -2\dfrac{d\left [ O_{2} \right ]}{dt}\ =\ \dfrac{1}{2}\dfrac{d\left [ N_{2}O_{5} \right ]}{dt}

Coefficients of the reactants and products are written as the reciprocals in the rate law expression.

It clearly means that rate of formation of N_{2}O_{5} is double of the rate of disappearance of oxygen.

Rate of formation of N_{2}O_{5} is half of the rate of disappearance of NO_2.

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Based on the three formulas shown, use one of them to solve for the purple yellow and red box and explain how you did it.
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P = 11.133 atm (purple)

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<h3>Further explanation  </h3>

Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated),:  

  • Boyle's law at constant T, P = 1 / V  
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So that the three laws can be combined into a single gas equation, the ideal gas equation  

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

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V = volume, liter  

n = number of moles  

R = gas constant = 0.08206 L.atm / mol K  

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To choose the formula used, we refer to the data provided

Because the data provided are temperature, pressure, volume and moles, than we use the formula PV = nRT

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V=8.3 L

P=1.8 atm

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\tt T=\dfrac{PV}{nR}\\\\T=\dfrac{1.8\times 8.3}{5\times 0.08205}\\\\T=36.42~K=-236.733^oC

  • Red box

T = 12 + 273.15 = 285.15 K

V=3.4 L

P=1.2 atm

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 3.4}{0.08205\times 285.15}\\\\n=0.174~mol

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