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const2013 [10]
3 years ago
10

The element chlorine (Cl) has two isotopes: chlorine-35 and chlorine-37. 75.5% of chlorine atoms have 18 neutrons and 17 protons

, and the other 24.5% have 20neutrons and 17protons. Using the isotopic composition provided, calculate the average atomic mass of chlorine.
Chemistry
1 answer:
Yuri [45]3 years ago
4 0

Answer:

35.8 u

Explanation:

The atomic mass of Cl is the <em>weighted average</em> of the atomic masses of its isotopes.

We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its <em>percent abundance</em>).

Atomic mass of Cl-35 = 17p + 18n = 17 × 1.007 u + 18 × 1.009 u

= 17.119 u + 18.162 u = 35.28 u

Atomic mass of Cl-37 = 17p + 20n = 17 × 1.007 u + 20 × 1.009 u

= 17.119 u + 20.180 u = 37.30 u

Set up a table for easy calculation.

0.755 × 35.28 u =  26.64  u

0.245 × 37.30 u =  <u>  9.138 u </u>

              TOTAL = 35.8     u

Note: The actual atomic mass of Cl is 35.45 u.

The calculated value above is incorrect because

(a) the given isotopic percentages are <em>incorrect</em> and

(b) the protons and neutrons have less mass when they are in the nucleus than when they are free. Thus, the calculated masses of Cl-35 and Cl-37 are <em>too high</em>.

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- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

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∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

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∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

  • Calculating ∆S°rxn:

∵  ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

  • Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

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