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CHM 111 Lab worksheets and Pre-lab

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These are documents from CHEM 111 from the university of Northern Virginia Community College. There is a wind spread of Molecular Geometry.









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Uploaded on
July 3, 2025
Number of pages
2
Written in
2024/2025
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Other
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Subjects

  • molecular geometry

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CHM 111 DETERMINE THE ENTHALPY OF CHEMICAL REACTION - HESS’S LAW

PRE-LAB

1. You will conduct the following three reactions in this experiment. In the space provided
below, write the balanced net ionic reaction equations from the descriptions. Use the
table of thermodynamic data in your text (or another approved resource) to calculate the
molar enthalpy of the reactions.
Reaction 1: An aqueous solution of sodium hydroxide reacts with an aqueous solution of
hydrochloric acid, yielding water.
Reaction 2: An aqueous solution of sodium hydroxide reacts with an aqueous solution of
ammonium chloride, yielding aqueous ammonium hydroxide, NH4OH, and water.
Reaction 3: An aqueous solution of hydrochloric acid reacts with aqueous ammonium
hydroxide, NH4OH, yielding aqueous ammonium chloride.
Reaction Balanced reaction equation ΔH (kJ/mol)

1 H+ (aq) + OH- (aq) → H2O (l) -55.90 kJ/mol

2 NaOH (aq) + NH4Cl (s) → NH4 (aq) + H2O (l) + NaCl (s) -3.70 kJ/mol

3 HCl (aq) + NH4OH → NH4Cl + H2O -52.30 kJ/mol


2. When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at
24.5°C, the temperature of the solution went up to 49.8°C.

a. Is this solution reaction exothermic or endothermic? Why?
- This solution is an exothermic reaction since the temperature of the
solution increases from 24.5 °C to 49.8°C. When there is an increase of
temperature then it is an exothermic reaction, which means heat or
energy is being released.

b. Calculate q of solution.
- Formula: Q = c x m x Δ T
- J of energy = 4.184 J
- Δ T = 49.8°C - 24.5°C = 25.3 K
- Q = 4.184 J x (5.12 g + 51.55 g) x 25.3K = 5995.81/ 1000 = 6.00 kJ
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