Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HNO3 in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the ΔH°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume 1) that no heat is lost to the calorimeter or the surroundings, and 2) that the density and the heat capacity of the resulting solution are the same as water. - 34.4 kJ/mol NaOH -169 kJ/mol NaOH -55.7 kJ/mol NaOH -27.9 kJ/mol NaOH -16.7 kJ/mol NaOH

Answer 1

Answer:

$\boxed{\text{-55.8 kJ/mol NaOH}}$

Explanation:

NaOH + HNO₃ ⟶ NaNO₃ + H₂O

There are two energy flows in this reaction.

$\begin{array}{cccl}\text{Heat from neutralization} & + &\text{Heat absorbed by water} & = 0\\q_{1} & + & q_{2} & =0\\n\DeltaH & + & mC\Delta T & =0\\\end{array}$

Data:

V(base) = 100.0 mL; c(base) = 0.300 mol·L⁻¹

V(acid) = 100.0 mL; c (acid) = 0.300 mol·L⁻¹  

        T₁ = 35.00 °C;          T₂ = 37.00 °C

Calculations:

(a) q₁

$n_{\text{NaOH}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}\\\\n_{\text{HNO}_{3}} = \text{0.1000 L } \times \dfrac{\text{0.300 mol}}{\text{1 L}} = \text{0.0300 mol}$

We have equimolar amounts of NaOH and HNO₃

n = 0.0300 mol

q₁ = 0.0300ΔH

(b) q₂

 V = 100.0 mL + 100.0 mL = 200.0 mL

m = 200.0 g

ΔT = T₂ - T₁ = 37.00 °C – 35.00 °C = 2.00 °C

q₂ = 200.0 × 4.184 × 2.00 = 1674 J

(c) ΔH

0.0300ΔH + 1674 = 0

          0.0300ΔH = -1674

                      ΔH = -1674/0.0300

                      ΔH = -55 800 J/mol

                      ΔH = -55.8 kJ/mol

$\Delta_{r}H^{\circ} = \boxed{\textbf{-55.8 kJ/mol NaOH}}$