A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 ml sample of 0.300 M HNO3 in a coffee cup calorimeter. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius. Determine the delta H (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HNO3. Assume no heat is lost to the calorimeter or the surroundings; and the density and heat capacity of the resulting solution are the same as water.
What is delta H?
0.1 L x 0.300 M NaOH = 0.0300 moles NaOH.
0.1 L x 0.300 M HNO3 = 0.0300 moles HNO3.
Mass water (200 mL) x specific heat water x (Tfinal-Tinitial) = q in joules. This is delta H.
delta H/mol is delta H from above divided by 0.03 mole.
Then you need to put it in kJ/mol for the question.
Ahh I had this right but calculated it wrong. Thanks a lot :)
Delta H is like a drama queen. It stands for the change in enthalpy during a chemical reaction. Enthalpy is a fancy word that represents the total energy content of a system. So basically, delta H tells us how much heat is either absorbed or released during a reaction. It's like measuring the emotional roller coaster of the reactants and products. Are they feeling hot or cold? Are they shedding tears of joy or sorrow? Delta H knows all the juicy details.
Delta H (ΔH) is the change in enthalpy, which represents the amount of heat absorbed or released during a chemical reaction at constant pressure. In other words, it is the heat transferred to or from the system during the reaction. It is usually measured in units of joules (J) or kilojoules (kJ), but in this case, we are asked to determine ΔH in units of kJ/mol of NaOH.
Delta H, or ΔH, refers to the change in enthalpy during a chemical reaction. Enthalpy is a measure of the total energy of a system, including both its internal energy and the work done by or on the system.
In this case, you are trying to determine the ΔH for the neutralization reaction between aqueous NaOH and HNO3. To do this, you can use the equation:
ΔH = q / n
Where:
- ΔH is the change in enthalpy in kJ/mol
- q is the heat transferred in the reaction, in Joules
- n is the number of moles of the limiting reactant involved in the reaction
To calculate q, the heat transferred, you can use the equation:
q = m * C * ΔT
Where:
- q is the heat transferred in Joules
- m is the mass of the solution in grams
- C is the specific heat capacity of the solution in J/(g·°C)
- ΔT is the change in temperature in °C
Since the density and heat capacity of the resulting solution are the same as water:
- The density of water is 1 g/mL
- The specific heat capacity of water is 4.18 J/(g·°C)
- The molar mass of NaOH is 39.997 g/mol
Now, let's calculate the values needed to find ΔH.
1. Calculate the mass and number of moles of NaOH:
- Mass = volume × density
Mass = 100.0 mL × 1 g/mL = 100.0 g
- Moles = mass / molar mass
Moles = 100.0 g / 39.997 g/mol = 2.500 mol
2. Calculate the heat transferred (q) using the formula:
- q = m × C × ΔT
- Convert the temperatures to Kelvin:
Initial temperature (T1) = 35.0 °C + 273.15 = 308.15 K
Final temperature (T2) = 37.0 °C + 273.15 = 310.15 K
- Calculate ΔT = T2 - T1
ΔT = 310.15 K - 308.15 K = 2.00 K
- Calculate q:
q = 100.0 g × 4.18 J/(g·°C) × 2.00 K = 836 J
3. Calculate ΔH using the formula:
- ΔH = q / n
- Convert J to kJ by dividing by 1000:
ΔH = 836 J / (2.500 mol × 1000) = 0.334 kJ/mol NaOH
Therefore, the ΔH for the neutralization reaction between NaOH and HNO3 is 0.334 kJ/mol NaOH.