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When \(1.00 \, \text{g}\) of solid \( \text{K} \) was added to water, about \(4610 \, \text{J}\) of energy was released.

What is the enthalpy of the reaction?

Sagot :

GinnyAnswer
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).
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