Question

An unknown base has a volume of 70 mL and a molarity of 2.00 and a specific heat capacity of 4.184 J/ g °C). The volume of oxalic acid is 80.00mL and a molarity of 1.80 and has a specific heat of 4.184 J/ g °C. The inital temperature was 20°C and the final temperature was 32.72 °C.Determine the ΔH of neutralization for the unknown base and state whether its exothermic or endothermic.

Answer 1

Sol :-

Number of moles of unkonwn base = Molarity x volume in L

= 2.00 M x 0.070 L

= 0.14 mol

Similarly,

Number of moles of oxalic acid= Molarity x volume in L

= 1.80 M x 0.080 L

= 0.144 mol

Now,

Amount of heat released (q) = Mass of the solution x specific heat capacity x change in temperature

=-(70 g + 80 g) x 4.184 J/ g °C x (32.72 - 20) °C

= -7983.072 J

So,

Enthalpy of neutralization = ΔH⁰ = q/Moles of limiting reactant

= -7983.072 J / 0.14 mol

= -57021.9 J/mol

= -57.02 KJ/mol

Hence, enthalpy of neutralization = ΔH0 = -57.02 KJ/mol

Because, sign of ΔH⁰ is negative, therefore this reaction is exothermic reaction.