In a coffee cup calorimeter, a vigourous reaction took place between potassium solid (0.258 g) and water. The total volume was 100 mL. If the temperature increased from 22.0 °C to 25.1 °C. Calculate the enthalpy change for the reaction in kJ/gram potassium. Assume the density of the solution after the reaction is 1 g/mL and the specific heat of 4.184 J/g-°C.

Question

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5027 J/g
1297 J/g
1.30 x103 J/g

Answer 1

The first step is to calculate the amount of heat released during the reaction using the equation:

q = mcΔT

Where q is the heat released, m is the mass of the solution (100 g), c is the specific heat of the solution (4.184 J/g-°C) and ΔT is the change in temperature (25.1 °C - 22.0 °C = 3.1 °C).

q = (100 g)(4.184 J/g-°C)(3.1 °C)
q = 1297.24 J

Next, we need to convert the amount of heat released to the enthalpy change for the reaction. Since we know the mass of potassium used, we can calculate the enthalpy change per gram of potassium using the equation:

ΔH = q/m

Where m is the mass of potassium used (0.258 g).

ΔH = 1297.24 J/0.258 g
ΔH = 5026.98 J/g

Rounding to three significant figures, we get:

ΔH = 5027 J/g

Therefore, the enthalpy change for the reaction between potassium and water is 5027 J/g of potassium.