Question

The weak acid, HX, is neutralized by sodium hydroxide as shown below: HX(aq) + NaOH(aq) → NaCN(aq) + H2O(1). Suppose you carried out this reaction in a coffee cup calorimeter using 50.0 mL of 0.500 M HX and 0.500 mL of 0.500 M NaOH. If the temperature of the calorimeter increases by 2.08 °C, what is the enthalpy of this neutralization reaction (in kJ/mol)? Assume the density of the solution is 1.00g/mL and that the specific heat is 4.184 J/g °C {Enter your value with the appropriate sign and value, but no units.} RUESTION 2 Colorless gas nitric oxide gas, NO, can react with oxygen to form the brown gas nitrogen dioxide, NO2, as shown below: 2 NO(g) + O2(g) → 2 NO2(g). Suppose you wanted to carry out this reaction in the laboratory. You could filled one gas syringe with 100. mL of nitric oxide and another with 110. mL of oxygen gas. When you connected the syringes and mixed the gases, the reaction mixture changes to a brown color indicating the the reaction had occurred. If the syringes were maintained at constant temperature and pressure, what would be the expected volume of the reaction mixture after the reaction? Hint 1: it is not simply 100+110 mL. Hint 2: nitric oxide is the limiting reagent, which means the end result is a combination of the product gas and the leftover oxygen gas.

Answer 1

Q1. enthalpy of neutralization = -34.8 kJ/mol

Q2. expected volume = 160 mL

Explanation

Q1. volume HX = 50.0 mL = 0.0500 L

concentration HX = 0.500 M

moles HX = (concentration HX) * (volume HX in Liter)

moles HX = (0.500 M) * (0.0500 L)

moles HX = 0.025 mol

Similarly, moles NaOH = 2.5 x 10^-4 mol

Total volume = (volume HX) + (volume NaOH)

Total volume = (50.0 mL) + (0.500 mL)

Total volume = 50.5 mL

total mass = (Total volume) * (density of solution)

total mass = (50.0 mL) * (1.00 g/mL)

total mass = 50.5 g

heat gained by solution = (total mass) * (specific heat of solution) * (temperature rise)

heat gained by solution = (50.5 g) * (4.184 J/g.^oC) * (2.08 ^oC)

heat gained by solution = 439.5 J

Heat lost by neutralization reaction = -(heat gained by solution)

Heat lost by neutralization reaction = -(439.5 J)

Heat lost by neutralization reaction = -439.5 J

enthalpy of neutralization = (Heat lost by neutralization reaction) / (moles NaOH)

enthalpy of neutralization = (-439.5 J) / (2.5 x 10^-4 mol)

enthalpy of neutralization = -1757949.44 J/mol

enthalpy of neutralization = -1758 kJ/mol