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GarryVolchara [31]
3 years ago
12

A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found

to depend on the initial concentration of A. Determine whether each statement is likely to be true or false for this reaction.
a. A doubling of the concentration of A doubles the rate of the reaction.
b. A plot of 1/[A] versus time is linear.
c. The half-life of the reaction gets longer as the initial concen- tration of A increases.
d. A plot of the concentration of A versus time has a constant slope.
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

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ololo11 [35]

Answer:

1) Maximun ammount of nitrogen gas: m_{N2}=10.682 g N_2

2) Limiting reagent: NO

3) Ammount of excess reagent: m_{N2}=4.274 g

Explanation:

<u>The reaction </u>

2 NO (g) + 2 H_2 (g) \longrightarrow N_2 (g) + 2 H_2O (g)

Moles of nitrogen monoxide

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n_{NO}=\frac{m_{NO}}{M_{NO}}

n_{NO}=\frac{22.9 g}{30 g/mol}=0.763 mol

Moles of hydrogen

Molecular weight: M_{H2}=2 g/mol

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1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted

m_{N2}=0.763 mol NO* \frac{1 mol N_2}{2 mol NO}*\frac{28 g N_2}{mol N_2}

m_{N2}=10.682 g N_2

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3) <u>Ammount of excess reagent</u>:

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8 0
3 years ago
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Answer:

a. Polar

b. Polar

c. Non-polar

d. Non-polar

Explanation:

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