A calorimeter contains 22.0 mL of water at 14.0 C. When 2.50 g of X (a substance with a molar mass of 66.0 g/mol) is added, it dissolves via the reaction X(s)+H_2O(l)--------> X(aq) and the temperature of the solution increases to 25.5 C. Calculate the enthalpy change, Delta H, for this reaction per mole of X. Assume that the specific heat and density of the resulting solution are equal to those of water [4.18 J/(g C) and 1.00 g/mL] and that no heat is lost to the calorimeter itself, nor to the surroundings.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
dH = delta H.
dH/g = q/g = q/2.50
dH/mol = q/mol = (q/2.50)*molar mass
1755.6 kj/g
To calculate the enthalpy change, delta H, for this reaction per mole of X, we need to use the equation:
delta H = q / n
where:
- delta H is the enthalpy change
- q is the heat absorbed or released by the reaction
- n is the number of moles of X involved in the reaction
First, we need to calculate the heat absorbed or released by the reaction, q. We can use the equation:
q = m * c * delta T
where:
- q is the heat absorbed or released by the reaction
- m is the mass of the water
- c is the specific heat capacity of water
- delta T is the change in temperature
Given information:
- Volume of water, V = 22.0 mL = 22.0 g (since 1 mL of water has a mass of 1 g)
- Specific heat capacity of water, c = 4.18 J/(g°C)
- Initial temperature, t1 = 14.0°C
- Final temperature, t2 = 25.5°C
First, calculate the mass of the water:
mass of water (m) = density * volume = 1.00 g/mL * 22.0 mL = 22.0 g
Next, calculate the change in temperature:
delta T = t2 - t1 = 25.5°C - 14.0°C = 11.5°C
Now we can calculate the heat absorbed or released by the reaction:
q = m * c * delta T = 22.0 g * 4.18 J/(g°C) * 11.5°C
Next, we need to calculate the number of moles of X involved in the reaction. We can use the equation:
moles of X = mass of X / molar mass of X
Given information:
- Mass of X, mX = 2.50 g
- Molar mass of X, MX = 66.0 g/mol
Calculate moles of X:
moles of X = 2.50 g / 66.0 g/mol
Finally, we can calculate the enthalpy change per mole of X:
delta H = q / n = (22.0 g * 4.18 J/(g°C) * 11.5°C) / (2.50 g / 66.0 g/mol)
Now, calculate delta H.
To calculate the enthalpy change, ΔH, for this reaction per mole of X, we need to use the equation:
ΔH = q / n
where ΔH is the enthalpy change, q is the heat exchanged, and n is the number of moles of X.
First, let's calculate the heat exchanged, q. We can use the equation:
q = m * C * ΔT
where q is the heat exchanged, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, the mass of water, m, can be calculated using the density of water:
m = V * ρ
where V is the volume of water and ρ is the density of water.
Given:
Volume of water, V = 22.0 mL
Density of water, ρ = 1.00 g/mL
m = 22.0 mL * 1.00 g/mL = 22.0 g
Next, we can calculate the change in temperature, ΔT:
ΔT = T_final - T_initial
where T_final is the final temperature and T_initial is the initial temperature.
Given:
Initial temperature, T_initial = 14.0 °C
Final temperature, T_final = 25.5 °C
ΔT = 25.5 °C - 14.0 °C = 11.5 °C
Now we can use the equation for q:
q = m * C * ΔT = 22.0 g * 4.18 J/(g°C) * 11.5 °C
q = 1072.84 J
Next, we need to calculate the number of moles of X. We can use the equation:
n = m / M
where n is the number of moles, m is the mass, and M is the molar mass.
Given:
Mass of X, m = 2.50 g
Molar mass of X, M = 66.0 g/mol
n = 2.50 g / 66.0 g/mol
n = 0.0379 mol
Finally, we can calculate the enthalpy change, ΔH:
ΔH = q / n = 1072.84 J / 0.0379 mol
ΔH = 28282.06 J/mol
Therefore, the enthalpy change, ΔH, for this reaction per mole of X is 28282.06 J/mol.