Abomb calorimeter, or a constant volume calorimeter, is a device often used to determine the heat of combustion of fuels and the energy content of foods. in an experiment, a 1.1070 g sample of fumaric acid (c4h4o4) is burned completely in a bomb calorimeter. the calorimeter is surrounded by 1.093ă—103 g of water. during the combustion the temperature increases from 22.10 to 24.52 â°c. the heat capacity of water is 4.184 j g-1â°c-1. the heat capacity of the calorimeter was determined in a previous experiment to be 891.1 j/â°c. assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of fumaric acid based on these data. c4h4o4(s) + 3o2(g) 2 h2o(l) + 4 co2(g) + energy

-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

it should be p1 v1 = p2 v2

a is the answer to this question

explanation: