In the reaction between formic acid (HCHO2) and sodium hydroxide, water and sodium formate (NaCHO2) are formed. To determine the heat of reaction, 75.0 mL of 1.07 M HCHO2 was placed in a coffee cup calorimeter at a temperature of 20.8 °C, and 45.0 mL of 1.78 M NaOH, also at 20.8 °C, was added. The mixture was stirred quickly with a thermometer, and its temperature rose to 25.3 °C. Write the balanced chemical equation for the reaction. Calculate the heat of reaction in joules. Assume that the specific heats of all solutions are 4.18 J g-1°C-1 and that all densities are 1.00 g mL-1. What is the heat of reaction per mole of acid (in units of kJ mol-1).
First, let's write the balanced chemical equation for the reaction:
HCHO2 + NaOH -> NaCHO2 + H2O
To calculate the heat of reaction, we can use the equation:
q = (m x C x ΔT)
Where:
q = heat (in joules)
m = mass (in grams)
C = specific heat (in J g-1°C-1)
ΔT = change in temperature (in °C)
First, let's calculate the mass of formic acid:
m(HCHO2) = volume x concentration x density
m(HCHO2) = 75.0 mL x 1.07 mol/L x 1.00 g/mL
m(HCHO2) = 80.25 g
Next, let's calculate the mass of sodium hydroxide:
m(NaOH) = volume x concentration x density
m(NaOH) = 45.0 mL x 1.78 mol/L x 1.00 g/mL
m(NaOH) = 80.10 g
Now, let's calculate the heat of reaction:
q(HCHO2) = m(HCHO2) x C x ΔT
q(HCHO2) = 80.25 g x 4.18 J g-1°C-1 x (25.3 °C - 20.8 °C)
q(HCHO2) = 1674.58 J
q(NaOH) = m(NaOH) x C x ΔT
q(NaOH) = 80.10 g x 4.18 J g-1°C-1 x (25.3 °C - 20.8 °C)
q(NaOH) = 1664.52 J
Now, let's calculate the total heat of reaction:
q(total) = q(HCHO2) + q(NaOH)
q(total) = 1674.58 J + 1664.52 J
q(total) = 3339.10 J
To convert the heat of reaction to kJ, divide by 1000:
q(total) = 3339.10 J / 1000
q(total) = 3.3391 kJ
Next, let's calculate the moles of formic acid:
n(HCHO2) = mass / molar mass
n(HCHO2) = 80.25 g / 46.03 g/mol
n(HCHO2) = 1.742 mol
Finally, let's calculate the heat of reaction per mole of acid:
ΔH per mole of acid = q(total) / n(HCHO2)
ΔH per mole of acid = 3.3391 kJ / 1.742 mol
ΔH per mole of acid = 1.916 kJ/mol
Therefore, the heat of reaction per mole of acid is 1.916 kJ/mol.
To write the balanced chemical equation for the reaction between formic acid (HCHO2) and sodium hydroxide (NaOH), we need to consider the stoichiometry of the reaction.
The general reaction between a carboxylic acid and a base, such as hydroxide, can be represented as follows:
acid (HCHO2) + base (NaOH) -> water (H2O) + salt (NaCHO2)
Based on this pattern, the balanced chemical equation for the reaction is:
HCHO2 + NaOH -> H2O + NaCHO2
Now, let's calculate the heat of reaction using the given information:
1. Calculate the moles of reactants used:
- HCHO2: moles = volume (L) × molarity (mol/L)
moles of HCHO2 = 0.075 L × 1.07 mol/L = 0.08025 mol
- NaOH: moles = volume (L) × molarity (mol/L)
moles of NaOH = 0.045 L × 1.78 mol/L = 0.0801 mol
2. Determine the limiting reactant:
The stoichiometric ratio between HCHO2 and NaOH is 1:1. Since the moles of HCHO2 and NaOH are very close, they are both in a 1:1 ratio.
3. Calculate the heat absorbed by the solution:
The heat absorbed (q) can be calculated using the equation:
q = mass (g) × specific heat (J/g°C) × ΔT (°C)
For the HCHO2 solution:
mass of HCHO2 = volume (mL) × density (g/mL)
= 75 mL × 1.00 g/mL = 75 g
ΔT = final temperature - initial temperature
= 25.3 °C - 20.8 °C = 4.5 °C
q(HCHO2) = 75 g × 4.18 J/g°C × 4.5 °C = 1419.75 J
For the NaOH solution:
mass of NaOH = volume (mL) × density (g/mL)
= 45 mL × 1.00 g/mL = 45 g
ΔT = final temperature - initial temperature
= 25.3 °C - 20.8 °C = 4.5 °C
q(NaOH) = 45 g × 4.18 J/g°C × 4.5 °C = 846.45 J
The total heat absorbed by the solution is given by:
q(total) = q(HCHO2) + q(NaOH)
= 1419.75 J + 846.45 J = 2266.2 J
4. Calculate the heat of reaction per mole of acid:
The heat of reaction per mole of acid can be calculated by dividing the total heat absorbed by the moles of HCHO2.
heat of reaction per mole of acid = q(total) / moles of HCHO2
= 2266.2 J / 0.08025 mol = 28217.32 J/mol
Finally, convert the heat of reaction from joules to kilojoules:
heat of reaction per mole of acid = 28217.32 J/mol = 28.22 kJ/mol
Therefore, the heat of reaction per mole of acid is 28.22 kJ/mol.
6.69 joules
NaOH + HCOOH ==> H2O + HCOONa
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
[note that mass H2O = mL HCOOH + mL NaOH since the density of H2O is 1.0 g/mL.]
delta H rxn/gram acid = q/grams HCOOH
dH rxn/mol acid = (q/g HCOOH)*molar mass HCOOH.