Zinc metal reacts with hydrochloric acid according to the following balanced equation: Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) When 0.101 g of Zn (s) is combined with enough HCl to make a 50.0 mL solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.27 °C to 23.65 °C. Calculate the enthalpy of reaction (ΔHrxn) in kJ/mol. Given: Assume that the density of the solution is 1.00 g/mL and the specific heat of the solution is 4.184 J/g·°C. IMPORTANT: Enter both the sign and the value (no units) using 3 significant figures
The Correct Answer and Explanation is:
To calculate the enthalpy of reaction (ΔHₓₓₓ) in kJ/mol, we first need to use the heat released in the reaction, and then determine how much energy was released per mole of zinc reacted.
Step 1: Calculate the heat released (q)
The formula for heat is:q=m×c×ΔTq = m \times c \times \Delta Tq=m×c×ΔT
Where:
- mmm = mass of the solution (g)
- ccc = specific heat of the solution (J/g·°C)
- ΔT\Delta TΔT = change in temperature (°C)
We are given:
- The mass of the solution: Since the density is 1.00 g/mL and the volume is 50.0 mL, the mass of the solution is equal to the volume in grams: m=50.0 gm = 50.0 \, \text{g}m=50.0g
- The specific heat of the solution: c=4.184 J/g\cdotp°Cc = 4.184 \, \text{J/g·°C}c=4.184J/g\cdotp°C
- The temperature change: ΔT=23.65∘C−22.27∘C=1.38∘C\Delta T = 23.65^\circ C – 22.27^\circ C = 1.38^\circ CΔT=23.65∘C−22.27∘C=1.38∘C
Now, calculating the heat released:q=50.0 g×4.184 J/g\cdotp°C×1.38∘C=288.76 Jq = 50.0 \, \text{g} \times 4.184 \, \text{J/g·°C} \times 1.38^\circ C = 288.76 \, \text{J}q=50.0g×4.184J/g\cdotp°C×1.38∘C=288.76J
Step 2: Calculate the moles of zinc reacted
The molar mass of zinc (Zn) is 65.38 g/mol. We are given the mass of zinc used:Mass of Zn=0.101 g\text{Mass of Zn} = 0.101 \, \text{g}Mass of Zn=0.101g
Now, calculate the moles of zinc reacted:Moles of Zn=Mass of ZnMolar mass of Zn=0.101 g65.38 g/mol=0.001544 mol\text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} = \frac{0.101 \, \text{g}}{65.38 \, \text{g/mol}} = 0.001544 \, \text{mol}Moles of Zn=Molar mass of ZnMass of Zn=65.38g/mol0.101g=0.001544mol
Step 3: Calculate the enthalpy change per mole of zinc
Now that we know the heat released (qqq) and the moles of zinc reacted, we can calculate the enthalpy of reaction per mole of zinc:ΔHrxn=qmoles of Zn\Delta H_{\text{rxn}} = \frac{q}{\text{moles of Zn}}ΔHrxn=moles of Znq
Substitute the values:ΔHrxn=288.76 J0.001544 mol=187,000 J/mol=187 kJ/mol\Delta H_{\text{rxn}} = \frac{288.76 \, \text{J}}{0.001544 \, \text{mol}} = 187,000 \, \text{J/mol} = 187 \, \text{kJ/mol}ΔHrxn=0.001544mol288.76J=187,000J/mol=187kJ/mol
Final Answer:
ΔHrxn=−187 kJ/mol\Delta H_{\text{rxn}} = -187 \, \text{kJ/mol}ΔHrxn=−187kJ/mol
The negative sign indicates that the reaction is exothermic (heat is released). So, the enthalpy of reaction is -187 kJ/mol.
