Answer:
-1960 kJ.
Step-by-step explanation:
C₄H₄O₄(s) + 3O₂(g) ⟶ 2H₂O((ℓ)) + 4CO₂(g) + Energy
There are three energy flows in this reaction.
From combustion + warm water + warm calorimeter = 0
q₁ + q₂ + q₃ = 0
nΔH + mCwΔT + CcalΔT = 0
Data:
Mass of fumaric acid = 1.1070 g
Mass of water = 1.093 × 10³ g
Cw = 4.184 J·°C⁻¹g⁻¹
T₁ = 21.10 °C
T₂ = 24.52 °C
Ccal = 891.1 J·°C⁻¹
Calculations:
(a) q₁
n = 1.1070 g × (1 mol/116.07 g)
n = 0.009 537 mol
q₁ = 0.009 537ΔH J
(b) q₂
ΔT = 24.52 – 21.10
ΔT = 3.42°C
q₂ = 1093 × 4.184 × 3.42
q₂ = 15 640 J
(c) q₃
q₃ = 891.1 × 3.42
q₃ =3048 J
(d) ΔH
0.009 537ΔH + 15 640 + 3048 = 0
0.009 537ΔH + 18 688 = 0
0.009 537ΔH = -18 688
ΔH = -18 688/0.009 537
ΔH = -1 959 413 J/mol
ΔH = -1960 kJ/mol
This is quite different from the actual value of -1334.70 kJ·mol⁻¹