Question

Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answer 1

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

$n = \frac{m}{M}$

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

$n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol$

The enthalpy change is:

$\Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ$

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!