First, we need to calculate the amount of H2 produced in the reaction using the mole ratio:
2 moles of Na produces 1 mole of H2
So, the amount of H2 produced is 0.5 moles (since we have 1 mole of Na)
Next, we need to calculate the amount of heat released or absorbed by the reaction, which can be calculated using the following formula:
q = m * c * ΔT
where q = heat, m = mass, c = specific heat, and ΔT = change in temperature.
In this case, we can use the total mass of the solution (sodium + water) and the specific heat of water since sodium has a negligible specific heat compared to water. The mass of the solution can be calculated as:
mass = volume * density = 1000 mL * 1.02 g/mL = 1020 g
ΔT = 65.0 °C - 20.0 °C = 45.0 °C
q = 1020 g * 4.00 J/g-°C * 45.0 °C = 183600 J
Finally, we can calculate the enthalpy change of the reaction using the following formula:
ΔH = q / n
where n = moles of H2 produced.
ΔH = 183600 J / 0.5 mol = 367200 J/mol
However, the units need to be converted from J/mol to kJ/mol:
ΔH = 367.2 kJ/mol
Therefore, the enthalpy change for the reaction in kJ/mol H2 is 367.2 kJ/mol.
In a constant pressure calorimeter, (cup calorimeter) a reaction occurred between sodium (23.00 g, 1.0mol) and water (1000 mL) based on the reaction shown below:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
The temperature of the resultant solution increases from 20.0 °C to 65.0 °C. If the volume of the solution remains as 1000 mL with a density of 1.02 g/mL and the specific heat changes of 4.00 J/g-°C. Calculate the enthalpy change for ther eaction in kJ/mol H2?
Hint: Use the mole ratio to determine the mol of H2 before calculating kJ/mol H2.
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