When 35.0 mL of 1.43 M NaOH at 22.0 degres Celsius is neutralized by 35.0 mL of HCL also at 22.0 degrees Celsius in a coffee-cup calorimeter, the temperature of the final solution rises to 31.29 degrees Celsius. Assume that the specific heat of all solutions i 4.18 J/ g* C, that the density of all solutions is 1.00 g/mL, and that volumes are additive.
A) calculate q for the reaction.
B) calculate q for the neautralization of one mole of NaOH.
i understand the calorimeter equation, but with that molarity in the equation, i got thrown off. help please.
A.
q = mass water x specific heat water x delta T = heat evolved by the neutralization of HCl with NaOH.
B. The above is q for how many moles? moles = L x M = 0.035L x 1.43 M = xx
Then q/xxmoles = joules/mole
Thank you!
To solve these questions, we first need to calculate the heat released or absorbed by the reaction (q). The formula to calculate q is:
q = mcΔT
where q is the heat exchanged, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
A) Calculate q for the reaction:
Since the volumes of NaOH and HCl used are the same and their densities are both 1.00 g/mL, the mass of the solution is equal to the volume.
m = 35.0 mL = 35.0 g
ΔT = 31.29°C - 22.0°C = 9.29°C
Using the specific heat capacity (c) of the solution (4.18 J/g*°C), we can now calculate q:
q = (35.0 g) * (4.18 J/g*°C) * (9.29°C)
q = 1351.08 J
Therefore, the heat released or absorbed by the reaction is approximately 1351.08 J.
B) Calculate q for the neutralization of one mole of NaOH:
To calculate q for the neutralization of one mole of NaOH, we need to find the number of moles of NaOH used. We know the volume and molarity of NaOH solution used, so we can use the formula:
moles = volume (in L) * molarity
volume = 35.0 mL = 0.035 L
molarity = 1.43 M
moles NaOH = 0.035 L * 1.43 mol/L
moles NaOH = 0.05005 mol
Now, we can calculate q:
q = (1351.08 J) / (0.05005 mol)
q = 27000 J/mol
Therefore, the heat released or absorbed by the neutralization of one mole of NaOH is approximately 27000 J/mol.
To solve this problem, we first need to understand the principles behind calorimetry and how to calculate the heat (q) absorbed or released in a reaction.
Calorimetry is the science of measuring the heat of reactions or physical changes. The equation used to calculate heat flow (q) in a reaction is:
q = m * c * ΔT
Where:
- q is the heat flow (in joules, J)
- m is the mass of the solution (in grams, g)
- c is the specific heat capacity of the solution (in J/g·°C)
- ΔT is the change in temperature (in °C)
A) Calculate q for the reaction:
In this problem, we are given the volume of the solutions instead of their masses. However, since the density of the solutions is given as 1.00 g/mL, we can assume that the mass is equal to the volume.
Given:
Volume of NaOH solution (V1) = 35.0 mL = 35.0 g
Volume of HCl solution (V2) = 35.0 mL = 35.0 g
Specific heat capacity (c) = 4.18 J/g·°C
Change in temperature (ΔT) = 31.29 °C - 22.0 °C = 9.29 °C
Using the equation q = m * c * ΔT, we can calculate the heat flow (q) for the reaction:
q = (35.0 g + 35.0 g) * 4.18 J/g·°C * 9.29 °C
q = 2 * 35.0 g * 4.18 J/g·°C * 9.29 °C
q = 2396.15 J
Therefore, the heat flow for this reaction is 2396.15 J.
B) Calculate q for the neutralization of one mole of NaOH:
To calculate the heat flow for the neutralization of one mole of NaOH, we need to determine the number of moles of NaOH used in the reaction.
Given:
Concentration of NaOH solution (c) = 1.43 M
Volume of NaOH solution (V1) = 35.0 mL = 35.0 g
Using the formula: moles = concentration * volume (in L), we can calculate the number of moles of NaOH used:
moles of NaOH = 1.43 M * (35.0 mL / 1000 mL/L)
moles of NaOH = 0.05005 mol
Now we can calculate the heat flow (q) for the neutralization of one mole of NaOH by dividing the total heat flow by the number of moles:
q = 2396.15 J / 0.05005 mol
q = 47952.55 J/mol
Therefore, the heat flow for the neutralization of one mole of NaOH is 47952.55 J/mol.