The reaction between 0.055 g of calcium with an excess of water was carried out in an ice calorimeter. The volume of water in the calorimeter decreased by 0.20 mL during the reaction. Calculate the enthalpy of the reaction (in kJ/mol).
To calculate the enthalpy of the reaction, we need to use the equation:
q = m * ΔT
where q is the heat released or absorbed during the reaction, m is the mass of the reactant (in this case, calcium), and ΔT is the change in temperature.
In this case, we're given the change in volume of water instead of the change in temperature, but we can use this information to calculate the heat absorbed by the water using the equation:
q = -ΔH * n
where ΔH is the enthalpy of the reaction and n is the number of moles of calcium.
First, we need to calculate the number of moles of calcium using its molar mass:
Molar mass of calcium = 40.08 g/mol
Number of moles of calcium = mass of calcium / molar mass of calcium
= 0.055 g / 40.08 g/mol
= 0.00137 mol
Next, we need to calculate the heat absorbed by the water:
q = -ΔH * n
Since the reaction is exothermic (heat released), q will be negative.
q = -(- C) * ΔT * mass of water
Since we're given the change in volume of water instead of the change in temperature, we need to convert the change in volume to the change in mass using the density of water:
density of water = 1 g/mL
change in mass of water = change in volume * density of water
= 0.20 mL * 1 g/mL
= 0.20 g
Now, we can substitute the values into the equation to get the value of q:
q = -(- C) * ΔT * mass of water
= C * ΔT * mass of water
Now, substitute the given values into the equation:
q = 0.00137 mol * ΔH
Now, rearrange the equation to solve for ΔH:
ΔH = q / (0.00137 mol)
Now, substitute the calculated value for q:
ΔH = (C * ΔT * mass of water) / (0.00137 mol)
Finally, substitute the given values into the equation and calculate ΔH:
ΔH = (C * ΔT * mass of water) / (0.00137 mol)
= (C * ΔT * 0.20 g) / (0.00137 mol)
= (C * ΔT * 0.20 g) / (0.00137 mol)
= C * ΔT * 145.99 kJ/mol
Therefore, the enthalpy of the reaction is C * ΔT * 145.99 kJ/mol.
To calculate the enthalpy of the reaction, we need to use the equation:
ΔH = q/n
where:
ΔH = enthalpy change of the reaction (in kJ/mol)
q = heat absorbed or released by the reaction (in kJ)
n = number of moles of the limiting reactant
First, let's determine the number of moles of calcium (Ca) in 0.055 g. The molar mass of calcium is 40.08 g/mol.
n(Ca) = mass(Ca) / molar mass(Ca)
n(Ca) = 0.055 g / 40.08 g/mol
n(Ca) ≈ 0.00137 mol
Since water is in excess, the number of moles of water can be considered as infinite, or at least much larger than the moles of calcium.
Next, we need to calculate the heat absorbed or released by the reaction (q). We can use the equation:
q = m × C × ΔT
where:
m = mass of water (in g)
C = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (in °C)
In this case, we are given the volume change of water (ΔV = -0.20 mL), so we can use the density of water to find the mass of water.
The density of water is approximately 1 g/mL.
mass(water) = volume(water) × density(water)
mass(water) = -0.20 mL × 1 g/mL
mass(water) = -0.20 g
Since the volume decreases, we need to use a negative sign in the calculation.
Now we need to determine the change in temperature (ΔT). However, the question does not provide this information, so we cannot directly calculate q.
To solve this, we need to know that the reaction was carried out in an ice calorimeter. This implies that the temperature remained constant at 0 °C during the reaction. Therefore, ΔT = 0 °C.
Now we can calculate q:
q = mass(water) × C × ΔT
q = -0.20 g × 4.18 J/g·°C × 0 °C
q = 0 J
Since there is no heat exchanged in this case (q = 0 J), the enthalpy change of the reaction can be calculated as:
ΔH = q / n
ΔH = 0 J / 0.00137 mol
ΔH = 0 J/mol
Therefore, the enthalpy of the reaction is 0 kJ/mol.