A 100 mL sample of 0.300M NaOH is mixed with a 100mL sample of 0.305M HNOs in a coffee cup calorimeter (constant pressure). If both solutions were initially at 35 C and the temperature of the resulting solution was recorded as 37°C, determine the Hrxn (in units of KJ/mol of the limiting reactant). Assume that no heat is lost to the calorimeter or the surroundings. Assume that all of the solutions have specific heat capacity and density of 4.186 /g.°C and 1 g/mL respectively.
To determine the enthalpy change (ΔHrxn) of the reaction in units of kJ/mol, we need to apply the equation:
ΔHrxn = q / n
where:
ΔHrxn is the enthalpy change of the reaction,
q is the heat transferred (in Joules),
n is the number of moles of the limiting reactant.
To solve this problem, we need to follow these steps:
1. Calculate the heat transferred (q) using the equation:
q = m * C * ΔT
where:
m is the mass of the solution in grams,
C is the specific heat capacity of the solution,
ΔT is the change in temperature.
To calculate m (mass of the solution), we need to consider the density of the solution:
m = volume * density
2. Determine the number of moles of limiting reactant (n).
To determine the limiting reactant, we need to compare the number of moles present in each solution. The reactant that has the smallest number of moles is the limiting reactant.
3. Substitute the values for q and n into the equation ΔHrxn = q / n and solve for ΔHrxn.
Let's go through each step using the provided information:
Step 1: Calculate the heat transferred (q)
q = m * C * ΔT
Given that the total volume of the solution is 100 mL + 100 mL = 200 mL = 200 g (since 1 mL of water is equal to 1 g),
and the change in temperature (ΔT) = final temperature - initial temperature = 37°C - 35°C = 2°C,
we can calculate q:
q = 200 g * 4.186 J/g.°C * 2°C
q = 1674.4 J
Step 2: Determine the number of moles of the limiting reactant (n)
To determine the limiting reactant, we need to calculate the moles of NaOH and HNO3 present in their respective solutions.
Moles NaOH = concentration * volume = 0.300 mol/L * 0.100 L = 0.030 mol
Moles HNO3 = concentration * volume = 0.305 mol/L * 0.100 L = 0.0305 mol
Since both have the same number of moles, we can conclude that neither NaOH nor HNO3 is in excess, and their ratio is 1:1. Thus, NaOH is the limiting reactant.
Step 3: Calculate ΔHrxn
ΔHrxn = q / n
ΔHrxn = 1674.4 J / 0.030 mol
ΔHrxn = 55,813 J/mol
To convert ΔHrxn into kJ/mol:
ΔHrxn = 55,813 J/mol * (1 kJ / 1000 J)
ΔHrxn = 55.813 kJ/mol
Therefore, the enthalpy change of the reaction (ΔHrxn) is 55.813 kJ/mol of the limiting reactant (NaOH).