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You have reacted 100.00mL of 1.353M aqueous sulfuric acid with 12.618g of sodium hydroxide solid. If all of the heat generated by this reaction is transferred to a 1.317kg block of copper metal initially at 16.82°C, what is the final temperature of the block of copper metal? (Specific heat of copper = 0.375J/g*°C)

Sagot :

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