In a coffee cup calorimeter 70.0 mL of 0.100 M AgNO3 and 70.0 mL of 0.100 M HCl are mixed to yield the following reaction.
Ag+(aq) + Cl ‾(aq) → AgCl(s)
If the two solutions are initially at 22.60°C, and if the final temperature is 23.39°C, calculate ΔH for the reaction in kJ/mol of AgCl formed. Assume a mass of 140.0 g for the combined solution and a specific heat capacity of 4.18 J °C-1 g-1.
And a density of 1.00 g/mL.
140 mL = 140 g liquid.
q = mass water x specific heat water x (Tfinal-Tinitial) = 0
You have mass and specific heat as well as Tf and Ti. solve for q. That is the heat generated. It is delta H, dH.
dH/0.007 mol gives dH/mol AgCl.
To calculate the enthalpy change (ΔH) for the reaction in kJ/mol of AgCl formed, we can use the equation:
ΔH = q / n
where:
- ΔH is the enthalpy change,
- q is the heat exchanged during the reaction, and
- n is the number of moles of AgCl formed.
First, let's calculate the heat exchanged (q) during the reaction using the equation:
q = m × c × ΔT
where:
- q is the heat exchanged,
- m is the mass of the combined solution,
- c is the specific heat capacity, and
- ΔT is the change in temperature.
Given:
- Mass of the combined solution = 140.0 g,
- Specific heat capacity (c) = 4.18 J °C^(-1) g^(-1),
- Initial temperature (T1) = 22.60°C, and
- Final temperature (T2) = 23.39°C.
First, calculate the change in temperature (ΔT):
ΔT = T2 - T1
ΔT = 23.39°C - 22.60°C
ΔT = 0.79°C
Next, calculate the heat exchanged (q):
q = m × c × ΔT
q = 140.0 g × 4.18 J °C^(-1) g^(-1) × 0.79°C
q = 466.076 J
To convert the heat to kJ, divide by 1000:
q = 466.076 J / 1000
q = 0.466076 kJ
Now, we need to find the number of moles of AgCl formed. Since the reaction is in a 1:1 ratio, the number of moles of AgCl will be the same as the number of moles of the limiting reactant. In this case, either AgNO3 or HCl can be the limiting reactant, so we need to determine which one is.
To do that, calculate the moles of AgNO3 and HCl:
Moles of AgNO3 = volume (in L) × molarity
Moles of AgNO3 = 70.0 mL × (1 L / 1000 mL) × 0.100 M
Moles of AgNO3 = 0.007 mol
Moles of HCl = volume (in L) × molarity
Moles of HCl = 70.0 mL × (1 L / 1000 mL) × 0.100 M
Moles of HCl = 0.007 mol
Since the moles of AgNO3 and HCl are equal, we know that they react in a 1:1 ratio. Therefore, the number of moles of AgCl formed (n) is also 0.007 mol.
Finally, calculate the enthalpy change (ΔH):
ΔH = q / n
ΔH = 0.466076 kJ / 0.007 mol
ΔH = 66.582286 kJ/mol
The enthalpy change (ΔH) for the reaction in kJ/mol of AgCl formed is approximately 66.6 kJ/mol.