The molar enthalpy of solution for ammonium chloride is 14.8kJ/mol. What is the final temperature observed when 20.0g of ammonium chloride is added to 125mL water at 20.0'C?
Ok thanks a lot for your help I understand it now :D
Can some help me with this question
To find the final temperature observed when ammonium chloride is added to water, we can use the equation:
\(q = m \cdot c \cdot \Delta T\)
where:
- \(q\) is the heat gained or lost by the system
- \(m\) is the mass of the substance
- \(c\) is the specific heat capacity of the substance
- \(\Delta T\) is the change in temperature
First, we need to calculate the heat gained or lost by the water using the molar enthalpy of solution. The molar enthalpy of solution (\(\Delta H_{sol}\)) is the heat change associated with dissolving one mole of the solute in a specific amount of solvent. It is given as 14.8 kJ/mol for ammonium chloride.
To convert this value to joules, we multiply it by 1000:
\(\Delta H_{sol} = 14.8 kJ/mol = 14.8 \times 1000 J/mol = 14800 J/mol\)
Next, we need to calculate the number of moles of ammonium chloride added to the water. We can use its molecular weight to do this.
The molecular weight of ammonium chloride (NH4Cl) is:
\(14.01 \, \text{g/mol}(N) + 1.01 \, \text{g/mol}(H) \times 4 + 35.45 \, \text{g/mol}(Cl) = 53.49 \, \text{g/mol}\)
Now, let's calculate the number of moles of ammonium chloride added to the water:
\(n = \frac{{\text{mass (g)}}}{{\text{molecular weight (g/mol)}}} = \frac{{20.0 \, \text{g}}}{{53.49 \, \text{g/mol}}} \approx 0.374 \, \text{mol}\)
Next, we need to find the heat gained or lost by the water. Since the heat gained by the water is equal to the heat lost by the ammonium chloride, we can set up an equation:
\(-q_{water} = q_{NH4Cl}\)
\(q_{water} = -\Delta H_{sol} \times n\)
\(q_{water} = -14800 \, \text{J/mol} \times 0.374 \, \text{mol} = -5535.2 \, \text{J}\)
Now, we can rearrange the heat equation to solve for the change in temperature (\(\Delta T\)):
\(q_{water} = m_{water} \times c_{water} \times \Delta T\)
From the given information, we have:
\(m_{water} = 125 \, \text{mL} = 125 \, \text{g}\)
\(c_{water} = 4.184 \, \text{J/g°C}\)
Substituting the values into the equation:
\(-5535.2 \, \text{J} = 125 \, \text{g} \times 4.184 \, \text{J/g°C} \times \Delta T\)
Simplifying the equation gives us:
\(\Delta T = \frac{{-5535.2 \, \text{J}}}{{125 \, \text{g} \times 4.184 \, \text{J/g°C}}} \approx -10.56 °C\)
The negative sign indicates a decrease in temperature. Therefore, the final temperature observed when 20.0 g of ammonium chloride is added to 125 mL of water at 20.0 °C is approximately 20.0 °C - 10.56 °C = 9.44 °C.
mols NH4Cl = 20 g/53.5 = about 0.37 but you need to do this and all other calculations here more accurately.
14.8 kJ/mol x 0.37 mol = 5.5 kJ = 5500 J.
Then -5500 J = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Solve for Tf.