Consider the following processes:
2ClF+O2-> Cl2O+F2O 167.4
2ClF3+2O2->Cl2O+3F2O 341.4
2F2+O2->2F2O –43.4
Calculate enthalpy for the reaction:
ClF3->ClF+F2
1. +217.5 kJ/mol
2. +130.2 kJ/mol
3. +108.7 kJ/mol
4. –217.5 kJ/mol
5. None of these
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To calculate the enthalpy for the reaction ClF3 -> ClF + F2, we can combine the given processes:
1. 2ClF + O2 -> Cl2O + F2O
2. 2ClF3 + 2O2 -> Cl2O + 3F2O
3. 2F2 + O2 -> 2F2O
To obtain the desired reaction, we can subtract the second and third processes from the first process.
2ClF + O2 - (2F2 + O2) -> ClF + F2
Now let's calculate the enthalpy change for each process and substitute them into the equation:
1. 2ClF + O2 -> Cl2O + F2O: 167.4 kJ/mol
2. 2ClF3 + 2O2 -> Cl2O + 3F2O: 341.4 kJ/mol
3. 2F2 + O2 -> 2F2O: -43.4 kJ/mol
Substituting the values:
167.4 - (341.4 + (-43.4)) = 167.4 - 341.4 + 43.4 = -130.6 kJ/mol
The enthalpy change for the reaction ClF3 -> ClF + F2 is approximately -130.6 kJ/mol.
Therefore, the correct answer is 2. +130.2 kJ/mol.
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