When 15.3g of NaNO3 was dissolved in water inside a constant pressure
calorimeter, the temperature fell from 25.00 °C to 21.56 °C. If the heat capacity of
the calorimeter (including the water it contains) is 1071 J/°C, What is the enthalpy
change for dissolving 1 mole of NaNO3? NaNO3 → Na+ + NO3- ∆H = ?
what formula do i use to solve this? i know q=C(change in T) but how does the 15.3g of nano3 affect the problem
You determine q for the system and that is q for 15.3 g NaNO3.
q/15.3 = J/gram. For 1 mole,
J/gram x (molar mass) = ??
To solve this problem, you can use the formula: q = C * ΔT, where q represents the heat absorbed or released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.
In this case, you have 15.3g of NaNO3, and you want to find the enthalpy change for dissolving 1 mole of NaNO3. To find the enthalpy change, we need to convert grams to moles.
Step 1: Calculate the number of moles of NaNO3:
First, determine the molar mass of NaNO3.
Na: atomic weight = 22.99 g/mol
N: atomic weight = 14.01 g/mol
O: atomic weight = 16.00 g/mol (x3 atoms)
So, the molar mass of NaNO3 = 22.99 + 14.01 + (16.00 * 3) = 85.00 g/mol
To calculate the number of moles:
moles = mass / molar mass
moles = 15.3 g / 85.00 g/mol
Step 2: Calculate the heat absorbed or released (q):
Using the formula q = C * ΔT, where C is the heat capacity of the calorimeter and ΔT is the change in temperature.
q = 1071 J/°C * (21.56 °C - 25.00 °C)
Step 3: Calculate the enthalpy change:
The enthalpy change (∆H) can be calculated using the equation:
∆H = q / moles
Substitute the values into the equation to find the enthalpy change (∆H).
To solve this problem, you can use the formula q = C * ΔT, where q is the heat energy transferred, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.
In this case, the heat energy transferred (q) represents the enthalpy change (∆H) for dissolving 1 mole of NaNO3. The heat capacity of the calorimeter (C) is given as 1071 J/°C.
Now, let's consider how the 15.3g of NaNO3 affects the problem. To determine the enthalpy change for dissolving 1 mole of NaNO3, we need to convert the mass of NaNO3 into moles. The molar mass of NaNO3 is 85 g/mol (22.99 g/mol for Na + 14.01 g/mol for N + 3 * 16.00 g/mol for O).
To calculate the number of moles, we use the formula:
moles = mass / molar mass
moles of NaNO3 = 15.3 g / 85 g/mol
moles of NaNO3 = 0.18 mol (rounded to two decimal places)
Since the question asks for the enthalpy change for dissolving 1 mole of NaNO3, we need to divide the heat energy transferred (q) by the number of moles of NaNO3 to obtain ∆H.
∆H = q / moles of NaNO3
Now, let's calculate q using the formula q = C * ΔT. In the problem, the temperature fell from 25.00 °C to 21.56 °C, so the change in temperature (ΔT) is 25.00 °C - 21.56 °C = 3.44 °C.
Now plug the values into the equation:
q = C * ΔT
q = 1071 J/°C * 3.44 °C
Calculating q:
q = 3680.24 J
Finally, we can calculate ∆H:
∆H = q / moles of NaNO3
∆H = 3680.24 J / 0.18 mol
Calculating ∆H:
∆H = 20,445.78 J/mol
Thus, the enthalpy change (∆H) for dissolving 1 mole of NaNO3 is approximately 20,445.78 J/mol.