A coffee-cup calorimeter is filled with 50 mL of 1 M NaOH and 50 mL of 1 M HCl, both at room temperature. After the reaction is complete, the final temperature of the solution is 32.5 degree C. Using the specific heat of water, calculate the heat of reaction, q (in Joules), and the Hrxn (in kJ/mol) for the neutralization of aqueous sodium hydroxide.
q = mcdT
q = 100*4.184*(Tfinal-Tinitial). You can fill in what room temperature is and calculate q in joules.
q = ? joules for 0.05 mols (i.e., 0.05L*1M) so scale up from 0.05 mols to 1 mol and convert to kJ/mol
Post your work if you get stuck.
Ah, chemistry! The subject that makes students go from "Oh, NaH!" to "Oh, NaCl!" in no time. Let's do some calculations and find those answers, shall we?
First, we need to find the temperature change (∆T). The final temperature (32.5°C) minus the initial room temperature should do the trick. But let's convert to Kelvin, just in case we end up needing that extra warmth. So, ∆T = (32.5 + 273.15) K - 298.15 K = 7.4 K.
Next, we need to find the heat transferred (q). Since the solution was in a coffee-cup calorimeter, we assume that the heat change is only due to the reaction. We'll use the equation q = m * c * ∆T, where m is the total mass of the solution and c is the specific heat of water (4.184 J/g•K).
The total mass of the solution is the sum of the masses of NaOH and HCl. Consider the density of water (which is approximately equal to the density of the solution) as 1 g/mL.
So, 50 mL of NaOH has a mass of 50 g, and 50 mL of HCl also has a mass of 50 g. The total mass of the solution is 50 g + 50 g = 100 g.
Plug these numbers into the equation: q = 100 g * 4.184 J/g•K * 7.4 K = 3089.6 J.
Lastly, we'll calculate ∆Hrxn using q and the molar amount of NaOH reacted.
First, determine the moles of NaOH in the solution. The concentration of NaOH is given as 1 M, meaning 1 mole of NaOH is dissolved in 1 liter (1000 mL) of solution. In this case, we have 50 mL of 1 M NaOH, so the moles of NaOH are 1 mole/L * 0.05 L = 0.05 moles.
Now, we divide q by the number of moles: ∆Hrxn = (q / moles) / 1000 = (3089.6 J / 0.05 moles) / 1000 = 61792 J/mol.
Finally, let's convert ∆Hrxn to kilojoules (kJ): ∆Hrxn = 61.792 kJ/mol.
So, the heat of reaction (q) is 3089.6 Joules, and the ∆Hrxn is 61.792 kJ/mol. I hope these calculations made your brain "brew-tifully heated" like a fresh cup of coffee!
To calculate the heat of reaction (q), you can use the equation:
q = m × c × ΔT
Where:
- q is the heat of reaction
- m is the mass of the solution (in grams)
- c is the specific heat capacity of water (4.18 J/g°C)
- ΔT is the change in temperature of the solution (in °C)
First, let's calculate the mass of the solution. Since we have 50 mL of NaOH and 50 mL of HCl, the total volume is 100 mL or 0.1 L.
The density of water is approximately 1 g/mL, so the mass of the solution is:
Mass = Volume × Density
Mass = 0.1 L × 1 g/mL
Mass = 0.1 kg
Next, we can calculate the change in temperature (ΔT). The initial temperature is room temperature, usually around 25°C. So, ΔT = Final temperature - Initial temperature.
ΔT = 32.5°C - 25°C
ΔT = 7.5°C
Now, let's substitute the values into the equation:
q = 0.1 kg × 4.18 J/g°C × 7.5°C
q = 3.135 J
The heat of reaction (q) is 3.135 Joules.
To calculate the ΔHrxn, we need to convert the heat of reaction from Joules to kilojoules and divide it by the number of moles of NaOH used.
Using the equation:
ΔHrxn = q / n
Where:
- ΔHrxn is the heat of reaction per mole of reactant (in kJ/mol)
- q is the heat of reaction (in J)
- n is the number of moles of reactant
To find the moles of NaOH, we can use the equation:
n = C × V
Where:
- n is the number of moles
- C is the concentration (in mol/L)
- V is the volume (in L)
Since the initial concentration of NaOH is 1 M, we have:
n = 1 mol/L × (50 mL / 1000 mL/L)
n = 0.05 mol
Now, let's convert the heat of reaction (q) to kilojoules:
q = 3.135 J = 0.003135 kJ
Finally, we can calculate ΔHrxn:
ΔHrxn = 0.003135 kJ / 0.05 mol
ΔHrxn = 0.0627 kJ/mol
Therefore, the heat of reaction (q) is 3.135 J, and the ΔHrxn is 0.0627 kJ/mol for the neutralization of aqueous sodium hydroxide.
To calculate the heat of reaction (q) and the ΔHrxn (standard enthalpy change) for the neutralization of aqueous sodium hydroxide (NaOH), we can use the equation:
q = m * c * ΔT
Where:
- q is the heat of reaction in Joules (J).
- m is the mass of the solution in grams (g).
- c is the specific heat of water, which is approximately 4.18 J/g°C.
- ΔT is the change in temperature (final temperature - initial temperature) in degrees Celsius (°C).
First, let's calculate the mass of the solution:
The solution is made up of 50 mL of 1 M NaOH and 50 mL of 1 M HCl. Since the molarity (M) is given in moles per liter (L), we can use this relationship to determine the moles in the 50 mL of the solution:
moles = molarity * volume (in liters)
moles of NaOH = 1 M * 0.050 L = 0.050 mol
moles of HCl = 1 M * 0.050 L = 0.050 mol
Since the concentration (M) is the same for both NaOH and HCl, they will react in a 1:1 ratio, resulting in the formation of water and a neutralized solution.
Next, let's calculate the mass of the solution:
The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol.
So, the mass of NaOH in the solution = moles of NaOH * molar mass of NaOH = 0.050 mol * 40.00 g/mol = 2.00 g.
Similarly, the mass of HCl in the solution is also 2.00 g.
Since we have equal volumes and concentrations of NaOH and HCl, the mass of the solution will be the sum of the masses of NaOH and HCl:
Mass of the solution = Mass of NaOH + Mass of HCl = 2.00 g + 2.00 g = 4.00 g.
Now, let's calculate the change in temperature:
ΔT = final temperature - initial temperature = 32.5 °C - room temperature.
Lastly, let's substitute the values into the heat of reaction equation (q = m * c * ΔT):
q = 4.00 g * 4.18 J/g°C * ΔT
Once you calculate the value of ΔT, you can use the equation ΔHrxn = q / moles of NaOH to find the ΔHrxn in kJ/mol.
Just divide the q value in Joules by the 0.050 moles of NaOH to get ΔHrxn in kJ/mol.