Consider the reaction HCl + NaOH ->NaCl + H2O
Given:
HCl Solution: 22 degrees celsius
NaOH Solution: 22 degrees celsius
Final Temperature: 26.1 degrees celsius
A. Calculate the amount of heat evolved when 15 mL of 1.0 M HCl was mixed with 35 mL of 1.0 M NaOH
B. Calculate Delta H per mole of water formed from his data
15 mL + 35 mL = 50 mL = 50 g assuming volumes are additive and density of the solution is the same as water (1.00 g/mL).
q = mass H2O x specific heat H2O x (Tfinal-Tintial).
That is q/50 g. Convert to moles for J/mol or to kJ/mol.
A. Oh boy, let's calculate the amount of heat evolved when HCl and NaOH get together for a little chemical reaction party! To do this, we can first find the moles of HCl and NaOH in the mixture:
Moles of HCl = Volume (L) x Molarity = 0.015 L x 1.0 M = 0.015 moles
Moles of NaOH = Volume (L) x Molarity = 0.035 L x 1.0 M = 0.035 moles
Now, since the reaction is a 1:1 ratio, we know that 0.015 moles of HCl will react with 0.015 moles of NaOH. Let's assume all the heat is evolved during the reaction (no heat loss or gain).
Next, we need to calculate the heat evolved using the equation:
q = m x c x ΔT
Where q is the heat evolved, m is the mass, c is the specific heat, and ΔT is the change in temperature. Since we know the final temperature is 26.1 degrees Celsius and assume the specific heat of the solution is the same as water (4.18 J/g°C), we can calculate the heat evolved.
But wait, we need the mass of the solution! To find that, we can use the density of the solution. Since I don't have the density information, I'll have to make a clownish guess. Let's say the density of the solution is 1 g/mL. That means the mass of the solution is:
Mass of Solution = Volume of Solution x Density of Solution
= (15 mL + 35 mL) x 1 g/mL
= 50 g
Now we can calculate the heat evolved:
q = m x c x ΔT
= 50 g x 4.18 J/g°C x (26.1°C - 22°C)
= 50 g x 4.18 J/g°C x 4.1°C
= 857.1 J
So, the amount of heat evolved when 15 mL of 1.0 M HCl was mixed with 35 mL of 1.0 M NaOH is approximately 857.1 joules. That's one hot reaction!
B. To calculate ΔH per mole of water formed, we need to know the moles of water formed in the reaction. Since HCl and NaOH react in a 1:1 ratio, we know that the moles of water formed is also 0.015 moles.
Now we can calculate ΔH per mole of water formed:
ΔH = q / n
= 857.1 J / 0.015 moles
≈ 57,140 J/mol
So, from this data, ΔH per mole of water formed is approximately 57,140 joules per mole. That's a lot of heat for just a little water!
To calculate the amount of heat evolved during the reaction, we can use the equation:
q = m * c * ΔT
where q is the heat evolved, m is the mass of the solution, c is the specific heat, and ΔT is the change in temperature.
A. Calculate the amount of heat evolved when 15 mL of 1.0 M HCl was mixed with 35 mL of 1.0 M NaOH:
Step 1: Calculate the initial and final temperatures
The initial temperature of both HCl and NaOH solutions is 22 degrees Celsius, and the final temperature is 26.1 degrees Celsius.
Step 2: Calculate the total mass of the solution
The total volume of the solution is the sum of the volumes of HCl and NaOH, which is 15 mL + 35 mL = 50 mL.
Density of water is approximately 1 g/mL, so 50 mL of the solution is equivalent to 50 g.
Step 3: Calculate the heat evolved
We need to find the change in temperature, ΔT = T_final - T_initial.
ΔT = 26.1 degrees Celsius - 22 degrees Celsius = 4.1 degrees Celsius.
The specific heat of a solution containing water is approximately 4.18 J/g°C.
Now we can calculate the amount of heat evolved:
q = m * c * ΔT
q = 50 g * 4.18 J/g°C * 4.1 degrees Celsius
q = 845.7 J
So, the amount of heat evolved when 15 mL of 1.0 M HCl was mixed with 35 mL of 1.0 M NaOH is approximately 845.7 J.
B. Calculate ΔH per mole of water formed from this data:
To calculate ΔH per mole of water formed, we need to convert the heat evolved (q) to moles of water.
Step 1: Calculate the number of moles of water formed from the balanced chemical equation.
The balanced chemical equation shows that 1 mole of water is formed when 1 mole of HCl reacts with 1 mole of NaOH.
Step 2: Calculate the moles of water formed from the volume and molarity of the solutions.
For HCl: 15 mL * 1.0 mol/L = 0.015 moles
For NaOH: 35 mL * 1.0 mol/L = 0.035 moles
Since the reactants are in a 1:1 stoichiometric ratio, the number of moles of water formed is equal to the limiting reactant, which is 0.015 moles.
Step 3: Calculate ΔH per mole of water formed.
ΔH = q / moles of water formed
ΔH = 845.7 J / 0.015 moles
ΔH = 56,380 J/mol
So, the value of ΔH per mole of water formed from this data is approximately 56,380 J/mol.