Consider the following reaction.
2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l) ΔH = -118 kJ
Calculate the heat when 114.8 mL of 0.500 M HCl is mixed with 300.0 mL of 0.400 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 414.8 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.
1. Determine the limiting reagent which will be either HCl or Ba(OH)2. It looks like HCl but you should confirm that before going further.
q = 118 kJ x (g HCl/2*molar mass HCl)
2. Then q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute and solve for Tf.
Well, this reaction seems to have a "hot" temper! Let's calculate the heat released first.
First, let's find the moles of HCl and Ba(OH)2:
Moles HCl = volume (L) × concentration (M) = (0.1148 L) × (0.500 mol/L) = 0.0574 mol
Moles Ba(OH)2 = volume (L) × concentration (M) = (0.3000 L) × (0.400 mol/L) = 0.1200 mol
Now, let's find the total moles of HCl and Ba(OH)2:
Total moles = minimum of (moles HCl, moles Ba(OH)2) = minimum of (0.0574 mol, 0.1200 mol) = 0.0574 mol
Since the stoichiometry of the reaction is 2:1 for HCl to Ba(OH)2, we'll divide the moles of HCl by 2 to find the moles of BaCl2 formed:
Moles BaCl2 = (0.0574 mol) / 2 = 0.0287 mol
Now, let's calculate the heat released using the equation:
Heat released = moles BaCl2 × ΔH
= (0.0287 mol) × (-118 kJ/mol)
= -3.3826 kJ
Alright, we've got the heat released, now let's move on to finding the final temperature of the mixture.
To do that, we can use the equation:
q = mcΔT
Where:
q is the heat gained or lost by the system (in this case, the mixture)
m is the mass of the mixture
c is the specific heat capacity of the mixture
ΔT is the change in temperature
We've already found that the heat released is -3.3826 kJ.
Plugging in the values, we get:
-3.3826 kJ = (414.8 g) × (4.18 J/°C·g) × (T_final - 25.0°C)
Now let's solve for T_final:
T_final - 25.0°C = -3.3826 kJ / [(414.8 g) × (4.18 J/°C·g)]
T_final - 25.0°C = -0.02 °C
Finally, let's add 25.0°C to both sides:
T_final = 25.0°C - 0.02°C
T_final = 24.98°C
So, the final temperature of the mixture would be approximately 24.98°C. Keep in mind that this calculation assumes no heat loss to the surroundings, as if the HCl and Ba(OH)2 react inside a perfectly insulated container.
To solve this problem, we can use the concept of heat transferred (q), which is given by the equation:
q = m * c * ΔT
Where:
q = heat transferred (in Joules)
m = mass of the solution (in grams)
c = specific heat capacity of the solution (in J/°C·g)
ΔT = change in temperature (in °C)
First, let's calculate the moles of HCl and Ba(OH)2 used in the reaction:
For HCl:
Volume = 114.8 mL = 0.1148 L
Concentration = 0.500 M
n(HCl) = Volume * Concentration
n(HCl) = 0.1148 L * 0.500 mol/L
n(HCl) = 0.0574 mol
For Ba(OH)2:
Volume = 300.0 mL = 0.3000 L
Concentration = 0.400 M
n(Ba(OH)2) = Volume * Concentration
n(Ba(OH)2) = 0.3000 L * 0.400 mol/L
n(Ba(OH)2) = 0.120 mol
From the balanced equation, we can see that the ratio of HCl to Ba(OH)2 is 2:1. Therefore, the limiting reagent is Ba(OH)2, which means that all of it will react and the amount of HCl used will be twice the moles of Ba(OH)2 used.
n(HCl) used = 2 * n(Ba(OH)2)
n(HCl) used = 2 * 0.120 mol
n(HCl) used = 0.240 mol
Next, let's calculate the heat released by the reaction using the given ΔH value:
q = ΔH * n(HCl) used
q = -118 kJ * 0.240 mol
Convert kJ to J:
1 kJ = 1000 J
q = -118,000 J * 0.240 mol
q = -28,320 J
Now, let's use the equation for heat transferred to calculate the temperature change:
q = m * c * ΔT
-28,320 J = 414.8 g * 4.18 J/°C·g * ΔT
ΔT = -28,320 J / (414.8 g * 4.18 J/°C·g)
ΔT = -16.99 °C
Finally, the final temperature of the mixture can be calculated by subtracting the temperature change from the initial temperature:
Final Temperature = Initial Temperature + ΔT
Final Temperature = 25.0°C - 16.99 °C
Final Temperature = 8.01 °C
Therefore, the final temperature of the mixture is approximately 8.01 °C.
To calculate the final temperature of the mixture, we need to use the principle of conservation of energy. The heat released by the exothermic reaction is absorbed by the mixture and causes a change in temperature.
First, let's calculate the heat released in the chemical reaction using the given equation:
-118 kJ for the complete reaction of 2 moles of HCl.
Now, let's determine the moles of HCl and Ba(OH)2 used in the reaction based on the given volumes and concentrations.
For HCl:
Volume = 114.8 mL
Concentration = 0.500 M
Calculate moles of HCl:
moles of HCl = Volume (L) x Concentration (mol/L) = 0.1148 L x 0.500 mol/L = 0.0574 mol
For Ba(OH)2:
Volume = 300.0 mL
Concentration = 0.400 M
Calculate moles of Ba(OH)2:
moles of Ba(OH)2 = Volume (L) x Concentration (mol/L) = 0.3000 L x 0.400 mol/L = 0.1200 mol
Since the balanced equation has a 1:2 ratio of moles of HCl to moles of Ba(OH)2, we can see that 0.0574 mol of HCl reacts with 0.0287 mol of Ba(OH)2.
To calculate the heat released in the reaction for the given amounts of reactants, we can use the ratio between the moles of the limiting reactant (which is the one in lesser amount) and the moles in the balanced equation. In this case, HCl is the limiting reactant.
Heat released in reaction = moles of HCl used x (heat released in the reaction / moles of HCl in balanced equation)
Heat released in reaction = 0.0574 mol x (-118 kJ / 2 mol) = -3.472 kJ
Now, let's calculate the heat absorbed by the mixture using the heat capacity equation:
Heat absorbed = mass x specific heat capacity x change in temperature
Mass of the mixture = 414.8 g
Specific heat capacity = 4.18 J/°C·g (needs to be converted to kJ/°C·g)
Change in temperature = final temperature - initial temperature
Since the mixture consists of water (2 H2O(l)) formed in the reaction, we can use the molar heat capacity of water (Cp = 75.3 J/mol·°C).
Heat released (in kJ) = Heat absorbed (in kJ)
-3.472 kJ = (414.8 g x 4.18 J/°C·g x (final temperature - 25.0°C)) / 1000 (to convert J to kJ)
Now, let's solve for the final temperature:
-3.472 kJ = (1732.024 J/°C x (final temperature - 25.0°C)) / 1000
-3.472 = 1.732024 (final temperature - 25)
(finally temperature - 25) = -3.472 / 1.732024
(final temperature - 25) = -2.005
final temperature = -2.005 + 25
final temperature = 22.995°C
Therefore, the final temperature of the mixture is approximately 22.995°C.