Assume that no heat is lost to the surroundings and calculate the enthalpy involved when 200. mL of .10 M potassium hydroxide and 200. mL of .10 M hydrochloric acid are mixed. Both solutions began at room temperature, 24.5 degrees celsius, and have densities very close to 1.00 g/mL. The highest temperature reached when the two solutions were mixed was 31.0 degrees celsius. Assume that the specific heat of the solution is equal to that of water, 4.184 J/g times K
To calculate the enthalpy change (ΔH) involved when the two solutions are mixed, we can use the equation:
ΔH = mcΔT
where:
m = mass of the solution
c = specific heat capacity of the solution
ΔT = change in temperature
Let's calculate the mass of the solution first:
The volume of the potassium hydroxide solution is 200 mL, with a density of 1.00 g/mL:
Mass of potassium hydroxide solution = volume × density = 200 mL × 1.00 g/mL = 200 g
The volume of the hydrochloric acid solution is also 200 mL, with a density of 1.00 g/mL:
Mass of hydrochloric acid solution = volume × density = 200 mL × 1.00 g/mL = 200 g
The total mass of the solution is the sum of the masses of the two solutions:
Total mass of the solution = mass of potassium hydroxide solution + mass of hydrochloric acid solution = 200 g + 200 g = 400 g
Now we can calculate the change in temperature (ΔT):
ΔT = highest temperature reached - initial temperature = 31.0°C - 24.5°C = 6.5°C
Since the specific heat capacity of the solution is the same as water (4.184 J/g°C), we can use this value for c.
Plugging the values into the equation, we get:
ΔH = (mass of solution) × (specific heat capacity) × (change in temperature)
= 400 g × 4.184 J/g°C × 6.5°C
Calculating the enthalpy change, we have:
ΔH = 10,336 J
Therefore, the enthalpy change when the two solutions are mixed is 10,336 J.
To calculate the enthalpy involved when two solutions are mixed, we can use the equation:
q = mcΔT
where q is the heat transferred, m is the mass of the solution, c is the specific heat capacity of the solution, and ΔT is the change in temperature.
First, we need to determine the mass of the solution. Since the density of the solutions is close to 1.00 g/mL, we can assume that the mass of each solution is equal to its volume (200 mL = 200 g). Therefore, the total mass of the solution is 400 g.
Next, we calculate the change in temperature (ΔT) by subtracting the initial temperature from the highest temperature reached:
ΔT = 31.0°C - 24.5°C = 6.5°C
Now, we can calculate the enthalpy change using the formula:
q = mcΔT
q = (400 g) x (4.184 J/g°C) x (6.5°C)
q = 108,806 J
So, the enthalpy change involved when the two solutions are mixed is 108,806 J.
You don't ask a question. I assume the problem is to determine the heat of neutralization in kJ/mol. Also, I assume there is no calorimeter constant so we ignore that.
Please check the post to make sure you have all of it correctly. The answer should be about 55 kJ/mol and it isn't coming out to that.