Question

Please answer asap!!

Given the change in the temperature from adding 3.00 grams of ammonium chloride to 100.00 mL of water, calculate the enthalpy of solution for ammonium chloride in units of kJ/mol.
kJ/mol
the change in temp was -1.99

Answer 1

`\large \boxed{\text{37.0 kJ$\cdot$mol$^(-1)$}}`

Step-by-step explanation:

There are two heat transfers involved in this problem.

Heat of solution of AlCl₃ + heat lost by water = 0

q₁ + q₂ = 0

nΔH + mCΔT = 0

Let's calculate the heats separately.

`\text{Moles of NH$_(4)$Cl}= \text{3.00 g NH$_(4)$Cl} * \frac{\text{1 mol NH$_(4)$Cl}}{\text{133.34 g NH$_(4)$Cl}} = \text{0.022 50 mol NH$_(4)$Cl}\\\\q_(1) = \text{0.022 50 mol} * \Delta H`

m = 100.00 g

C = 4.18 J·°C⁻¹g⁻¹

ΔT = -1.99 °C

q₂ = mCΔT = 100.00 g × 4.184 J·°C⁻¹g⁻¹ × (-1.99 °C) = -832.6 J

`q_(1) + q_(2) = \text{0.022 50 mol} * \Delta H - \text{832.6 J} = 0\\\text{0.022 50 mol} * \Delta H = \text{832.6 J}\\\Delta H = \frac{\text{832.6 J}}{\text{0.022 50 mol}} = \text{37 010 J/mol} =\textbf{37.0 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\\\text{The heat of solution of NH$_(4)$Cl is $\large \boxed{\textbf{37.0 kJ$\cdot$mol$^{\mathbf{-1}}$}}$}`