Question

When 0.625 g Mn is combined with enough hydrochloric acid to make 100.0 mL of solution in a coffee-cup calorimeter, all of the Mn reacts, raising the temperature of the solution from 23.5 °C to 28.8 °C. Find ΔHrxn for the reaction as written. (Assume that the specific heat capacity of the

Answer 1

Answer:

$\boxed{\text{-194 kJ/mol}}$

Explanation:

Mn + 2HCl ⟶ MnCl₂ + H₂

There are two energy flows in this reaction.

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

Data:

Mass of Mn = 0.625 g  

                V = 100.0 mL

               T₁ = 23.5 °C

               T₂ = 28.8 °C

Calculations:

(a) Moles of Mn

$n = \text{0.625 g Mn} \times \dfrac{\text{1 mol Mn }}{\text{54.94 g Mn}} = \text{0.011 38 mol Mn}$

(b) Mass of solution

$m = \text{100.0 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100.0 g}$

(c) ΔT

ΔT = T₂ - T₁ = 28.8 °C – 23.5 °C = 5.3 °C

(d) q₁

$q_{1} = \text{0.011 38 mol Mn} \times \Delta_{r}H = 0.01138 \Delta_{r}H \text{ mol}$

(e) q₂

q₂ = 100.0 × 4.184 × 5.3  = 2220 J

(f) ΔH

$\begin{array}{rcl}0.01138 \Delta_{r}H + 2220 & = & 0\\0.01138 \Delta_{r}H & = & -2220\\\\\Delta_{r}H & = & \dfrac{-2220}{0.01138}\\\\ & = & \text{-194000 J/mol}\\ & = & \boxed{\textbf{-194 kJ/mol}}\\\end{array}$