Answer:
70.137 °C
Explanation:
The reaction generated from the question can be expressed as:
[tex]2NaOH_{(s)} + H_2SO_{4(aq)} \to Na_2SO_{4(aq)} +2H_2O_{(l)}[/tex]
The enthalpy reaction:
[tex]\Delta H^0 _{rxn} = \sum \Delta H^0 _{f (products)} - \sum \Delta H^0 _{f (reactants)}[/tex]
[tex]\Delta H^0 _{rxn} =[2 \times \Delta H^0 _f (H_2O) + \Delta H^0 _f (Na_2SO_4) ] -[2 \Delta \times H^0 _f (NaOH) + \Delta H^0 _f (H_2SO_4) ][/tex]
Repacing the values of each compound at standard enthalpy conditions;
[tex]\Delta H^0 _{rxn} =[2 \times -279.4 + (-1384.49)]-[(2\times -418) -913]\ kJ[/tex]
[tex]\Delta H^0 _{rxn} =-194.29 \ kJ[/tex]
no of moles of NaOH = 12.618g/39.99 g/mol
= 0.3155 mol
no of moles of H₂SO₄ = molarity of H₂SO₄ × Volume
= 1.3553 mol/L × 100 × 10⁻³ L
= 0.13553 mol
From the reaction,
1 mol of NaOH = 2 × mol of H₂SO₄
Since mol of NaOH is greater than that of H₂SO₄, then NaOH is the excess reagent and H₂SO₄ is the limiting reactant
∴
1 mol of H₂SO₄ yields = - 194.29 kJ
0.13553 mol of H₂SO₄ will yield;
[tex]= \dfrac{-194.29 \ kJ}{1 \ mol} \times 0.13553 \ mol[/tex]
= -26.332124 kJ
= -26332.12 J
Finally,
Heat(q) = [tex]m_{(copper)} \times C_{(copper)}\times \Delta T[/tex]
26332.12 J = 1.317 × 10³ g × 0.375 J/g°C ×ΔT
26332.12 J = 493.875 J/° C × ΔT
26332.12 / 493.875 = ΔT
ΔT = 53.317 °C
[tex]T_f - T_i = 53.317 ^0 C[/tex]
[tex]T_f[/tex]- 16.82 °C = 53.317 °C
[tex]T_f[/tex] = (53.317 + 16.82) °C
[tex]T_f[/tex] = 70.137 °C
