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The enthalpy change of formation of Mn(NO₃)₂(s) is -696 kJ mol⁻¹ - CIE - A-Level Chemistry - Question 9 - 2014 - Paper 1
Question 9
The enthalpy change of formation of Mn(NO₃)₂(s) is -696 kJ mol⁻¹. The enthalpy change of formation of MnO₂(s) is -520 kJ mol⁻¹. The enthalpy change of formation of N... show full transcript
Worked Solution & Example Answer:The enthalpy change of formation of Mn(NO₃)₂(s) is -696 kJ mol⁻¹ - CIE - A-Level Chemistry - Question 9 - 2014 - Paper 1
Step 1
Calculate the enthalpy change for the reaction
Answer
To find the standard enthalpy change of the reaction, we can use the formula:
In this reaction:
- Reactants: Mn(NO₃)₂(s)
- Products: MnO₂(s) and 2NO₂(g)
Thus, the enthalpy change is calculated as follows:
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Enthalpy of formation for the reactant:
- Mn(NO₃)₂(s): -696 kJ/mol
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Enthalpy of formation for the products:
- MnO₂(s): -520 kJ/mol
- NO₂(g): 2 × 33 kJ/mol = 66 kJ/mol
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Summing these:
- Total for products = -520 kJ/mol + 66 kJ/mol = -454 kJ/mol
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Now plug in these values into the formula:
Thus, the standard enthalpy change for this reaction is +242 kJ/mol.
Step 2