Calculate the enthalpy change (ΔH) for: HCl(aq) + NH3(aq) → NH4Cl(aq)
The reaction of 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NH3 causes a temperature rise of 6.4 ˚C in the resulting 100.0 mL of solution. The NH4Cl solution has a density of 1.005 g/mL and a specific heat of 3.97 Jg-1˚C-1.
(My book says ΔH should be reported with no digits beyond the first uncertain digit, which I don't understand?)
My Work:
1) q=(100.5g)*(3.97Jg^-1C^-1)*(6.4C)=-2553.504J
2) ΔH=-2553.504J/.050moles *1 mole NH3=-51070.08J/1molrxn*1 kj/1000J=-510kj/molrxn
Did I do it right? What should I use for digits for ΔH based on what my book says? Thanks again so much!
Yes, you did it right. The answer should be reported as -510 kJ/mol.
Well, it seems like you're almost there! Let me check your work and provide some clarification on reporting the ΔH value.
1) The calculation for q looks good:
q = (100.5 g) * (3.97 J/g°C) * (6.4°C) = -2553.504 J
2) However, you made a small mistake in the next step. The moles of NH3 should be determined first, not HCl:
moles NH3 = (50.0 mL) * (1.00 mol/L) * (0.0500 L / 1000 mL) = 0.00250 mol NH3
ΔH = (-2553.504 J) / (0.00250 mol) = -1021416.32 J/mol
Now, about the reporting of ΔH. When your book says "ΔH should be reported with no digits beyond the first uncertain digit," it means you should round your value to one decimal place. In this case, the first uncertain digit is the 4 after the decimal point. So, your final answer for ΔH would be -1021416.3 J/mol.
And just to clarify, if you choose to convert J to kJ, you should divide by 1000, not multiply, so the final answer in kJ/mol would be -1021.4163 kJ/mol.
Hope that clears things up!
Your calculations are mostly correct. However, to calculate the enthalpy change (ΔH) for the reaction, you need to consider the balanced equation first:
HCl(aq) + NH3(aq) → NH4Cl(aq)
From the balanced equation, you can see that the reaction involves the formation of 1 mole of NH4Cl. Therefore, the moles of NH4Cl formed will be equal to the moles of HCl or NH3 that react.
To calculate the moles of NH4Cl formed:
moles NH4Cl = (volume NH4Cl solution) x (density NH4Cl solution) / (molar mass NH4Cl)
Given that the volume of the NH4Cl solution is 100.0 mL and the density is 1.005 g/mL, and the molar mass of NH4Cl is 53.49 g/mol:
moles NH4Cl = (100.0 mL) x (1.005 g/mL) / (53.49 g/mol) = 1.8835 mol
Since the reaction stoichiometry is 1 mole NH4Cl : 1 mole HCl or NH3, the moles of HCl or NH3 is also 1.8835 mol.
Now, let's calculate the heat transferred in the reaction:
q = (mass solution) x (specific heat solution) x (temperature change)
The mass of the solution is the sum of the masses of HCl and NH3:
mass solution = (volume HCl) x (density HCl) + (volume NH3) x (density NH3)
= (50.0 mL) x (1.00 g/mL) + (50.0 mL) x (1.00 g/mL)
= 100.0 g
Using the obtained values, the heat transferred (q) is:
q = (100.0 g) x (3.97 J/g·°C) x (6.4 °C) = 2553.6 J (rounded to four significant figures)
Now, let's calculate the enthalpy change (ΔH):
ΔH = q / (moles HCl or NH3)
= 2553.6 J / 1.8835 mol
= 1355 J/mol
The answer should be reported with no digits beyond the first uncertain digit, which in this case is the "5" in 1355 J/mol.
To calculate the enthalpy change (ΔH) for the given reaction, you have followed the correct steps. However, there is one minor error in your calculation.
Here's the correct approach:
1) Calculate the heat absorbed or released by the reaction using the formula q = mcΔT, where q is the heat, m is the mass of the solution, c is the specific heat, and ΔT is the temperature change.
Given:
- Volume of HCl = 50.0 mL
- Concentration of HCl = 1.00 M
- Volume of NH3 = 50.0 mL
- Concentration of NH3 = 1.00 M
- Temperature change = 6.4 ˚C
- Density of NH4Cl solution = 1.005 g/mL
- Specific heat of NH4Cl solution = 3.97 Jg^-1˚C^-1
First, calculate the moles of HCl and NH3 used:
- Moles of HCl = (50.0 mL * 1.00 mol/L) / 1000 mL = 0.050 mol
- Moles of NH3 = (50.0 mL * 1.00 mol/L) / 1000 mL = 0.050 mol
Next, calculate the mass of the NH4Cl solution formed:
- Mass of NH4Cl solution = (100.0 mL * 1.005 g/mL) = 100.5 g
Then, calculate the heat absorbed or released by the reaction:
- q = (mass of the solution) * (specific heat) * (temperature change)
- q = (100.5 g) * (3.97 Jg^-1˚C^-1) * (6.4 °C) = 2551.136 J or -2551.136 J (because the reaction is exothermic)
Finally, calculate the ΔH for the reaction:
- ΔH = q / (moles of NH3)
- ΔH = (-2551.136 J) / (0.050 mol) = -51022.72 J/mol
The correct answer for ΔH is -510 kJ/mol (reported with no digits beyond the first uncertain digit, as per your book's instructions).
Therefore, your initial calculation of -510 kJ/mol is correct. Just remember to round your answer to the correct significant digits as per your book's instructions.