27.0 mL of 1.00 M HClO4 are mixed with 27.0 mL of 1.00 M LiOH in a coffee cup calorimeter. The calorimeter constant is 28.1 J/C. Both were initially at 21.31 C. Calculate delta H(rxn) in kJ/mol for the reaction.
4.180*50*20.07=4191.65
To calculate the ΔH(rxn) in kJ/mol for the reaction, we need to use the formula:
ΔH(rxn) = q / n
where q is the heat transferred (in J), and n is the number of moles of the limiting reactant.
Step 1: Calculate the heat transferred (q)
We can use the equation:
q = m * C * ΔT
where m is the mass of the solution (in g), C is the specific heat capacity of the solution (in J/g°C), and ΔT is the change in temperature (in °C).
Since we know the volumes and concentrations of the reactants, we need to calculate the mass of the solution.
Step 1.1: Calculate the mass of the solution
The molar mass of HClO4 is 100.46 g/mol. Using the volume and concentration of HClO4:
n(HClO4) = V(HClO4) * C(HClO4)
= 27.0 mL * (1.00 mol/L) * (1 L / 1000 mL)
= 0.027 mol
mass(HClO4) = n(HClO4) * Molar mass(HClO4)
= 0.027 mol * 100.46 g/mol
= 2.7174 g
Similarly, we calculate the mass of LiOH:
n(LiOH) = V(LiOH) * C(LiOH)
= 27.0 mL * (1.00 mol/L) * (1 L / 1000 mL)
= 0.027 mol
Molar mass of LiOH = 6.94 + 16.00 + 1.01 = 23.95 g/mol
mass(LiOH) = n(LiOH) * Molar mass(LiOH)
= 0.027 mol * 23.95 g/mol
= 0.6477 g
The total mass of the solution is:
mass(solution) = mass(HClO4) + mass(LiOH)
= 2.7174 g + 0.6477 g
= 3.3651 g
Step 1.2: Calculate ΔT
The initial and final temperatures are both 21.31 °C, so ΔT = 0.
Step 2: Calculate q
Since ΔT = 0, q = 0.
Step 3: Calculate n (moles of the limiting reactant)
The stoichiometry of the reaction is 1:1 between HClO4 and LiOH. Therefore, the limiting reactant is HClO4.
n = n(HClO4) = 0.027 mol
Step 4: Calculate ΔH(rxn)
ΔH(rxn) = q / n
= 0 / 0.027 mol
= 0 kJ/mol
Therefore, the ΔH(rxn) for the reaction is 0 kJ/mol.
To calculate the change in enthalpy (ΔH) for a reaction, we can use the equation:
ΔH = q / n
Where:
ΔH is the change in enthalpy (in J/mol)
q is the heat transferred (in J)
n is the number of moles of the limiting reactant involved in the reaction
To calculate the heat transferred (q), we can use the equation:
q = C × ΔT
Where:
q is the heat transferred (in J)
C is the calorimeter constant (in J/°C)
ΔT is the change in temperature (in °C)
First, let's calculate the q value for the reaction using the equation above:
ΔT = Tf - Ti
Where:
Tf is the final temperature of the mixture in the coffee cup calorimeter
Ti is the initial temperature of the mixture in the coffee cup calorimeter
In this case, the initial temperature is 21.31 °C, and to find the final temperature, we need to assume that the heat exchange between the reactants and the calorimeter is complete, so the final temperature is the same as the temperature of the mixture.
Now, let's calculate the final temperature:
Tf = 21.31 °C
ΔT = Tf - Ti
ΔT = 21.31 °C - 21.31 °C
ΔT = 0 °C
Since there is no change in temperature, the heat transferred (q) is equal to zero.
Now, let's calculate the number of moles of the limiting reactant involved in the reaction. To determine the limiting reactant, we need to compare the moles of HClO4 and LiOH.
First, let's convert the volumes of the solutions to moles using the given molarities:
Molarity (M) = moles / volume (L)
27.0 mL of 1.00 M HClO4:
moles of HClO4 = 1.00 M × 0.0270 L
moles of HClO4 = 0.0270 mol
27.0 mL of 1.00 M LiOH:
moles of LiOH = 1.00 M × 0.0270 L
moles of LiOH = 0.0270 mol
The balanced chemical equation for the reaction between HClO4 and LiOH is:
HClO4 + LiOH → LiClO4 + H2O
From the balanced equation, we can see that the stoichiometric ratio between HClO4 and LiOH is 1:1, meaning that 1 mole of HClO4 reacts with 1 mole of LiOH.
Since the number of moles of HClO4 and LiOH are the same (0.0270 mol each), we can conclude that 0.0270 mol is the number of moles of the limiting reactant.
Now, let's calculate ΔH by substituting the values into the equation:
ΔH = q / n
ΔH = 0 J / 0.0270 mol
ΔH = 0 J/mol
Therefore, the value of ΔH is 0 J/mol.
Note: It seems that the reaction did not produce or absorb any heat in this case. Double-check if the reaction is supposed to be exothermic or endothermic.