Question

Write the balanced complete ionic and net ionic equations for Reaction A, B and C.

For reaction A calculate the heat of neutralization, in kJ / mol, for reaction B and C calculate the heat of reaction, in kJ / mol..

1. Reaction A : Aqueous sodium hydroxide, a strong electrolyte, reacts with aqueous hydrochloric acid, a strong electrolyte.

Contains: 25.0 mL 1.0 M HCl and  25.0 mL 1.0 M NaOH

Ti:25.0 C Tf: 31.67 C

2. Reaction B: Solid sodium hydroxide, a strong electrolyte, dissolves in aqueous hydrochloric acid.

Contains: 50.0 mL 0.50 M HCl and 1 gram sodium hydroxide solid

Ti: 25.0 C Tf: 36.97 C

3. Reaction C: Solid sodium hydroxide, a strong electrolyte, dissolves in water to form an aqueous solution.

Contains: 50.0 mL Distilled H₂O and  1 gram sodium hydroxide solid

Ti: 25.0 C Tf: 30.30 C

Answer 1

1) For reaction A, first we need to find q of reaction. It is given by formula

q = m*c*T

T = T_f - T_i

T = 31.67-25.0 = 6.67^oC

a) First we need to find mass of solution and for this find moles of reactant and then convert it in mass

Convert voume in L.

25.0ml*(1L/1000 mL) = 0.025 L

Molarity = moles*L

1.0 moles/L = moles/0.025L

moles = 0.025 L*1.0 mole/L = 0.025 moles of HCl and NaOH

b) Convert this moles in grams

0.025 mol NaOH*(40.0g NaOH/1 mol NaOH) = 1.0g NaOH

0.025 mol HCl*(36.46g HCl/1 mol HCl) = 0.9115g HCl

Hence total mass will be 1.0+0.9115 = 1.9115g

c) Use this mass, change in temperature and c=4.184J/g^oC in formula

q = 1.9115 g *4.184 J/g^oC *6.67^oC

= 53.3448 J

Since q(reaction) = -q(solution)

Hence q(reaction) = -53.3448 J

d) Convert this in kJ. Hence -53.3448 J*(1kJ/1000 J) = -0.0533448 kJ

H_rxn = q(reaction)/n

Where n-number of moles

e) H_rxn = -0.0533448 kJ/0.025 mol

= -2.1338 kJ/mol = -2.13 kJ/mol

Hence heat of neutralization for given reaction will be -2.13 kJ/mol.

In multiple question we solve only one question at a time. Thanks.