When a solution containing 8.00 g NaOH in 50.0 g of water at 25 °C is added to a solution of 8.00 g of HCl in 250.0 g of water at 25 °C in a calorimeter, the temperature of the solution increases to 33.5 °C. Assuming that the specific heat heat of the solution is 4.184 J/(g۰°C) and the calorimeter absorbs a negligible amount of heat, calculate ∆H in kJ for the reaction
NaOH(aq) + HCl(aq) →NaCl(aq) + H2O(l)
8g NaOH = 8/40 = 0.2 mols
8g HCl = 8/36.5 = 0.22 mol
So NaOH is the limiting reagent.
q = mass x secific heat H2O x (Tfinal-Tinitial)
q = (50+250) x 4.184 x (33.5-25) = ?
q is delta H in J (convert to kJ) and that is all the post asks for. Usually these problems want kJ/mol
When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.0 °C is added to asolution of 8.00 g of HCl in 250.0 g of water at 25.0 °C in a calorimeter, the temperature ofthe solution increases to 33.5 °C. Assuming that the specific heat of the solution is 4.184Jg ∙ °C and that of the calorimeter absorbs a negligible amount of heat, calculate ∆ H in kJfor the reactionNaOH(aq)+ HCl(aq) → NaCl(aq)+ H2 O(
Well, it seems like we have a little chemistry puzzle here. Don't worry, I shall solve it for you with a touch of humor!
To find the enthalpy change (∆H) for the reaction, we can use the formula:
q = m x C x ΔT
where:
q is the heat absorbed or released by the reaction,
m is the mass of the solution,
C is the specific heat capacity of the solution,
and ΔT is the change in temperature.
First, let's calculate the heat absorbed by the NaOH solution:
q1 = m1 x C x ΔT
m1 = 8.00 g NaOH + 50.0 g water = 58.0 g
q1 = 58.0 g x 4.184 J/(g۰°C) x (33.5 °C - 25 °C)
Now, let's calculate the heat absorbed by the HCl solution:
q2 = m2 x C x ΔT
m2 = 8.00 g HCl + 250.0 g water = 258.0 g
q2 = 258.0 g x 4.184 J/(g۰°C) x (33.5 °C - 25 °C)
Finally, let's calculate the total heat absorbed by the reaction:
q total = q1 + q2
Now, since we want the enthalpy change (∆H) in kJ, we can convert the heat to kilojoules:
∆H = q total / 1000
So, substitute the values and crunch the numbers, my friend! Good luck!
To calculate ∆H for the reaction, we can use the equation:
∆H = q / n
Where:
∆H is the heat of the reaction in kJ
q is the heat absorbed or released in J
n is the amount of substance in moles
First, we need to calculate the heat (q) absorbed or released by the reaction. This can be done using the equation:
q = m × c × ∆T
Where:
m is the mass of the solution in grams
c is the specific heat of the solution in J/(g°C)
∆T is the change in temperature in °C
In this case, the mass of the solution (m) is the sum of the masses of water and NaOH or HCl.
m = mass of water + mass of NaOH or HCl
Calculating the mass of the water:
mass of water = 50.0 g + 250.0 g = 300.0 g
Calculating the mass of NaOH:
mass of NaOH = 8.00 g
Calculating the mass of HCl:
mass of HCl = 8.00 g
Now, we can calculate the heat (q) absorbed or released by the reaction using the equation:
q = m × c × ∆T
∆T = final temperature - initial temperature
∆T = 33.5°C - 25.0°C = 8.5°C
Let's calculate the heat (q) absorbed or released by the reaction:
q = (300.0 g + 8.00 g + 8.00 g) × 4.184 J/(g°C) × 8.5°C
q = 31216.8 J
Next, we need to calculate the number of moles (n) of reactants or products involved in the reaction. Since the reaction is 1:1 between NaOH and HCl in terms of moles, we can use either one to calculate the moles. Let's use the moles of NaOH:
n = mass / molar mass
The molar mass of NaOH is:
Na = 22.990 g/mol
O = 16.000 g/mol
H = 1.0078 g/mol
molar mass of NaOH = 22.990 g/mol + 16.000 g/mol + 1.0078 g/mol = 39.9978 g/mol (approximately 40 g/mol)
n = 8.00 g / 40 g/mol = 0.20 mol
Finally, we can calculate ∆H by dividing q by the number of moles:
∆H = q / n
∆H = 31216.8 J / 0.20 mol = 156,084 J/mol = 156.084 kJ/mol
Therefore, the value of ∆H for the reaction is 156.084 kJ.