Answer :
To determine the enthalpy of the reaction when 1.00 g of solid potassium (K) is added to water, follow these steps:
1. Identify the mass and energy released:
- Mass of potassium (K): 1.00 grams.
- Energy released: 4610 joules.
2. Determine the molar mass of potassium (K):
- The molar mass of potassium (K) is approximately 39.1 grams per mole (g/mol).
3. Calculate the moles of K used in the reaction:
- Moles of K = (mass of K) / (molar mass of K)
- Moles of K = 1.00 g / 39.1 g/mol
- Moles of K ≈ 0.0256 moles (more specifically, approximately 0.025575 moles).
4. Calculate the enthalpy of the reaction:
- The enthalpy of the reaction, ΔH, is given by the energy released per mole of potassium.
- Enthalpy of the reaction = Energy released / Moles of K
- Enthalpy of the reaction = 4610 J / 0.025575 moles
- Enthalpy of the reaction ≈ 180251 J/mol (more precisely, approximately 180251.0 J/mol).
Thus, the enthalpy of the reaction is approximately 180251 joules per mole (J/mol).
1. Identify the mass and energy released:
- Mass of potassium (K): 1.00 grams.
- Energy released: 4610 joules.
2. Determine the molar mass of potassium (K):
- The molar mass of potassium (K) is approximately 39.1 grams per mole (g/mol).
3. Calculate the moles of K used in the reaction:
- Moles of K = (mass of K) / (molar mass of K)
- Moles of K = 1.00 g / 39.1 g/mol
- Moles of K ≈ 0.0256 moles (more specifically, approximately 0.025575 moles).
4. Calculate the enthalpy of the reaction:
- The enthalpy of the reaction, ΔH, is given by the energy released per mole of potassium.
- Enthalpy of the reaction = Energy released / Moles of K
- Enthalpy of the reaction = 4610 J / 0.025575 moles
- Enthalpy of the reaction ≈ 180251 J/mol (more precisely, approximately 180251.0 J/mol).
Thus, the enthalpy of the reaction is approximately 180251 joules per mole (J/mol).