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Novay_Z [31]
4 years ago
9

An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the

course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically.
The reaction of compound A forming compound B was studied and the following data were collected:

Time (s) [A](M)
0. 0.184
200. 0.129
500. 0.069
800. 0.031
1200. 0.019
1500. 0.016

a.) What is the average reaction rate between 0. and 1500. s?
b.) What is the average reaction rate between 200. s and 1200. s?
c.) What is the instantaneous rate of the reaction at t=800 s?
Chemistry
1 answer:
Sauron [17]4 years ago
8 0

Answer:

a) 0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b) 0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c) Instantaneous rate of the reaction at t=800 s :

Instantaneous rate  : \frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s

Explanation:

Average rate of the reaction is given as;

R_{avg}=-\frac{\Delta A}{\Delta t}=\frac{A_2-A_1}{t_2-t_1}

a.) The average reaction rate between 0.0 s and 1500.0 s:

At 0.0 seconds the concentration was  = A_1=0.184 M

t_1=0.0s

At 1500.0 seconds the concentration was  = A_2=0.016 M

t_2=1500 s

R_{avg]=-\frac{0.016 M-0.184 M}{1500.0 s-0.0 s}=0.000112 M/s

0.000112 M/s is the average reaction rate between 0.0 seconds and 1500.0 seconds.

b.) The average reaction rate between 200.0 s and 1200.0 s:

At 0.0 seconds the concentration was  = A_1=0.129 M

t_1 =200.0 s

At 1500.0 seconds the concentration was  = A_2=0.019M

t_2=1200 s

R_{avg]=-\frac{0.019 M-0.129M}{1200.0s-200.0s}=0.00011 M/s

0.00011 M/s is the average reaction rate between 200.0 seconds and 1200.0 seconds.

c.) Instantaneous rate of the reaction at t=800 s :

At 800 seconds the concentration was  = A=0.031 M

t =800.0 s

Instantaneous rate  : \frac{0.031 M}{800.0 s}=3.875\times 10^{-5} M/s

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Answer:

NaH₂PO₄ =  1.876 g

Na₂HPO₄ =  4.879 g

Combinations: H₃PO₄ and Na₂HPO₄; H₃PO₄ and Na₃HPO₄

Explanation:

To have a buffer at 7.540, the acid must be in it second ionization, because the buffer capacity is pKa ± 1. So, we must use pKa2 = 7.198

The relation bewteen the acid and its conjugated base (ion), is given by the Henderson–Hasselbalch equation:

pH = pKa + log[A⁻]/[HA], where [A⁻] is the concentration of the conjugated base, and [HA] the concentration of the acid. Then:

7.540 = 7.198 + log[A⁻]/[HA]

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[A⁻]/[HA] = 10^{0.342}

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[A⁻] = 2.198*[HA]

The concentration of the acid and it's conjugated base must be equal to the concentration of the buffer 0.0500 M, so:

[A⁻] + [HA] = 0.0500

2.198*[HA] + [HA] = 0.0500

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NaH₂PO₄ + Na₂HPO₄ → HPO₄⁻² + 3Na + H₂PO₄⁻

The second ionization is:

H₂PO₄⁻ ⇄ HPO₄⁻² + H⁺

So, H₂PO₄⁻ is the acid form, and its concentration is the same as NaH₂PO₄, and HPO₄⁻² is the conjugated base, and its concentration is the same as Na₂HPO₄ (stoichiometry is 1:1 for both).

So, the number of moles of these salts are:

NaH₂PO₄ = 0.01563 M * 1.000 L = 0.01563 mol

Na₂HPO₄ = 0.03436 M* 1.000 L = 0.03436 mol

The molar masses are, Na: 23 g/mol, H: 1 g/mol, P: 31 g/mol, and O = 16 g/mol, so:

NaH₂PO₄ = 23 + 2*1 + 31 + 4*16 = 120 g/mol

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Na₂HPO₄ = 0.03436 mol * 142 g/mol = 4.879 g

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H₃PO₄ and Na₃HPO₄ (same reason).

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