The reaction between 0.045 g of calcium with an excess of water was carried out in an ice calorimeter as used in this lab. The volume of water in the calorimeter decreased by 0.18 mL during the reaction
a) Write the equation for the reaction which occurs. Is it a redox reaction?
b) Calculate the enthalpy of the reaction (in kJ/mol). Show all steps of calculation.
I have part a. It is a redox reaction. but part b, i have no idea on how to solve enthalpy change!
Please Help!
If you mean the volume of ice decreased by .18ml, you can calculate the energy used to melt that amount of ice.
Then normalize that energy to joules/mole or joules/gram.
To calculate the enthalpy change of a reaction, you can use the equation:
ΔH = q / n
where ΔH is the enthalpy change (in kJ/mol), q is the heat absorbed or released by the reaction (in kJ), and n is the number of moles of the substance undergoing the reaction.
To determine the heat change (q), we can use the equation:
q = mCΔT
where q is the heat change (in kJ), m is the mass of water (in g), C is the specific heat capacity of water (4.18 J/g°C or 4.18 J/gK), and ΔT is the change in temperature (in °C or K).
First, we need to find the number of moles of calcium (Ca) reacted. Using the atomic mass of calcium (40.08 g/mol), the mass of calcium (0.045 g), and Avogadro's number (6.022 x 10^23 mol^-1), we can calculate the number of moles (n) as follows:
n = m / M
n = 0.045 g / 40.08 g/mol
= 0.0011 mol
Now, we need to find the heat change (q). Given that the volume of water decreased by 0.18 mL, we can calculate the mass of water (m) as follows:
m = V x ρ
where m is the mass of water (in g), V is the volume of water (in mL), and ρ is the density of water (1 g/mL).
m = 0.18 mL x 1 g/mL
= 0.18 g
Next, we need to find the change in temperature (ΔT) of the water. Since the reaction occurred in an ice calorimeter, we can assume that the initial and final temperatures of the water are 0°C. Therefore, ΔT = 0 - 0 = 0°C.
Now, we can calculate the heat change (q):
q = mCΔT
q = 0.18 g x 4.18 J/g°C x 0°C
= 0 J
Since the heat change (q) is 0 J, this means that no heat was absorbed or released by the reaction.
Finally, we can calculate the enthalpy change (ΔH) using the equation:
ΔH = q / n
ΔH = 0 J / 0.0011 mol
= 0 J/mol
Therefore, the enthalpy change of the reaction is 0 kJ/mol.