Consider the reaction:
2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l)
∆H = - 118 kJ
Calculate the heat when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.50 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25 oC and the final mixture has a mass of 400.0 g and a specific heat 4.18 J/g oC, calculate the final temperature of the mixture.
2H^+ + 2OH^- ==> 2H2O
When 2 mol H^+ react with 2 mols OH^-, the heat produced is 118 kJ. Therefore it it 118/2 kJ/mol and we have 0.05 mols H^+ = 2.95 kJ.
2,950 J = 400 g x specific heat x (Tf-25)
Solve for Tf.
To calculate the heat when the two solutions are mixed, we can use the formula:
q = m * C * ΔT
Where:
q = heat (in joules)
m = mass of the solution (in grams)
C = specific heat capacity of the solution (in J/g oC)
ΔT = change in temperature (in oC)
1. Calculate the mass of each solution:
For 100.0 mL of 0.500 M HCl:
Molarity (M) = mol/L
0.500 M = (mol HCl) / (0.100 L)
mol HCl = 0.500 * 0.100 = 0.050 mol
Molar mass of HCl = 1.01 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
Mass of HCl = (0.050 mol) * (36.46 g/mol) = 1.823 g
For 300.0 mL of 0.50 M Ba(OH)2:
Molarity (M) = mol/L
0.50 M = (mol Ba(OH)2) / (0.300 L)
mol Ba(OH)2 = 0.50 * 0.300 = 0.150 mol
Molar mass of Ba(OH)2 = 137.33 g/mol (Ba) + 2 * (1.01 g/mol (H) + 16.00 g/mol (O)) = 171.33 g/mol
Mass of Ba(OH)2 = (0.150 mol) * (171.33 g/mol) = 25.70 g
2. Calculate the total mass of the final mixture:
Mass of final mixture = 1.823 g + 25.70 g = 27.52 g
3. Calculate the initial heat of the reaction:
ΔH = -118 kJ = -118,000 J
4. Calculate the heat released or absorbed in the reaction:
q = -ΔH = -(-118,000 J) = 118,000 J
5. Calculate the change in temperature of the mixture:
q = m * C * ΔT
118,000 J = (27.52 g) * (4.18 J/g oC) * ΔT
ΔT = 118,000 J / (27.52 g * 4.18 J/g oC) = 1030.52 oC
6. Calculate the final temperature of the mixture:
Final temperature = Initial temperature + ΔT = 25 oC + 1030.52 oC = 1055.52 oC
Therefore, the final temperature of the mixture is approximately 1055.52 oC.
To calculate the heat when two solutions are mixed, we can use the equation:
q = mC∆T
where q is the heat, m is the mass, C is the specific heat, and ∆T is the change in temperature.
First, let's calculate the moles of HCl and Ba(OH)2 in the solutions.
n(HCl) = (100.0 mL) * (0.500 mol/L) = 0.0500 mol
n(Ba(OH)2) = (300.0 mL) * (0.500 mol/L) = 0.1500 mol
Now, let's calculate the limiting reagent. The stoichiometry of the reaction tells us that the balanced equation requires 2 moles of HCl for every 1 mole of Ba(OH)2. Therefore, we need to compare the moles of HCl and Ba(OH)2 to determine the excess and limiting reagent.
Since HCl has 0.0500 mol and Ba(OH)2 has 0.1500 mol, HCl is the limiting reagent.
Next, let's calculate the heat released using the relationship:
q = ∆H * n(HCl)
q = (-118 kJ) * (0.0500 mol) = -5.90 kJ
Now, let's calculate the mass of the final mixture:
mass = (100.0 mL + 300.0 mL) * (1.00 g/mL) = 400.0 g
Using the equation:
q = mC∆T
we can rearrange it to solve for ∆T:
∆T = q / (mC)
∆T = (-5.90 kJ) / (400.0 g * 4.18 J/g oC) = -0.0035 oC
Therefore, the final temperature of the mixture would be approximately 25 oC + (-0.0035 oC) = 24.9965 oC, or approximately 25 oC (rounded to three decimal places).