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Chlorine exists naturally as a mixture of two isotopes, \(^{35}Cl\) and \(^{37}Cl\), in the abundance ratio of 3:1 - CIE - A-Level Chemistry - Question 1 - 2014 - Paper 1
Question 1
Chlorine exists naturally as a mixture of two isotopes, \(^{35}Cl\) and \(^{37}Cl\), in the abundance ratio of 3:1. The mass spectrum of chlorine consists of five pe... show full transcript
Worked Solution & Example Answer:Chlorine exists naturally as a mixture of two isotopes, \(^{35}Cl\) and \(^{37}Cl\), in the abundance ratio of 3:1 - CIE - A-Level Chemistry - Question 1 - 2014 - Paper 1
Step 1
Suggest the mass numbers for these five peaks and the identities of the species responsible.
Answer
The five peaks in the mass spectrum of chlorine correspond to the following mass numbers and species:
- Mass number 35, species (^{35}Cl)
- Mass number 37, species (^{37}Cl)
- Mass number 70, species (^{35}Cl_2)
- Mass number 72, species (^{35}Cl^{+}Cl)
- Mass number 74, species (^{37}Cl^{+}Cl)
Thus, the peaks indicate the presence of both isotopes and their respective molecular forms.
Step 2
Predict the ratios of the abundances of the three species with the highest mass numbers.
Answer
The two isotopes of chlorine have an abundance ratio of 3:1. The species with the highest mass numbers are:
- (^{37}Cl) (mass number 37)
- (^{35}Cl^{+}Cl) (mass number 70)
- (^{37}Cl^{+}Cl) (mass number 74)
Given the ratios, the predicted ratios of the abundances among these top three species will be:
- Ratio of (^{35}Cl/^{37}Cl^{+}Cl) is 3:1
- Ratio of (^{37}Cl/^{35}Cl^{+}Cl) is 1:1.5
- Ratio of (^{37}Cl^{+}Cl/^{35}Cl^{+}Cl) is 1:3
Therefore, the final ratio for the three species is 9:6:1.
Step 3
Use the following data, together with relevant data from the Data Booklet, to calculate a value for the lattice energy of strontium chloride.
Answer
To calculate the lattice energy of strontium chloride, we can use the Born-Haber cycle. The relevant values are:
[ \Delta H_f^{\circ}(SrCl_2) = \text{-830 kJ mol}^{-1} ] [ \Delta H_{atom}(Sr) = +164 kJ mol^{-1} ] [ \Delta H_{EA}^{(Cl)} = -349 kJ mol^{-1} ]
The cycle can be set up as:
[ \Delta H_f = \Delta H_{atom}(Sr) + 2 \Delta H_{EA}(Cl) + \text{lattice energy} ]
Rearranging:
[ \text{lattice energy} = \Delta H_f - \Delta H_{atom}(Sr) - 2 \Delta H_{EA}(Cl) ]
Substituting the numbers:
[ \text{lattice energy} = -830 - 164 - 2(-349) ] [ = -830 - 164 + 698 ] [ = -296 kJ mol^{-1} ]
Thus, the calculated lattice energy of strontium chloride is -296 kJ mol^{-1}.