Temperature increases by 1.126 degree C. Calculate the heat given off by the bur

Question

Temperature increases by 1.126 degree C. Calculate the heat given off by the burning magnesium, in kJ g^-1 and in kJ mol^-1. The enthalpy of combustion of benzoic acid (C_6H_5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be -3226.7 kJ mol^-1. When 0.9862 g of benzoic acid was oxidized, the temperature rose from 21.84 degree C to 25.67 degree C. What is the heat capacity of the calorimeter? In a separate experiment, 0.4654 g of x-D-glucose (C_6H_12O_6) was oxidized in the same calorimeter, and the temperature rose from 21.22 degree C to 22.28 degree C. Calculate the enthalpy of combustion of glucose, the value of _r U for the combustion, and the molar enthalpy of formation of glucose. HCl is mixed with 2.00 Times 10^2 mL of 0.431 M Ba(OH)_2

Explanation / Answer

C6H5COOH

Hcomb = -3226.7 kJ/mol

a)

m = 0.9862 g of Benzoic Acid

T1 = 21.84

T2 = 25.67

find C of calorimeter

Q= C*dT

C = Q/dT

Frist, change the mass to moles

mol = mass/MW = 0.9862/122.12 = 0.0081 mol of B.Ac

1 mol --> -3226.7 kJ

0.0081 mol --> 26.06 kJ

Now, find C

C = Q/dT = 26.06 kJ/ (25.67-21.84) = 6.80 kJ/°C

b)

m = 0.4654 g of Glucose C6H12O6

Same Calorimeter, so C = 6.80 kJ/°C

T1 = 21.22

T2 = 22.28

find Hcombustion, dU, molar enthalpy of formaiton of glucose

Q = C*dT

Q = 6.80 kJ/°C * (22.28-21.22C) = 7.21 kJ for those 0.4654 g

change to mol

MW of glucose = 180

mol = mass/MW = 0.4654/180 = 0.002585 mol

Hrxn = 7.21 kJ/0.002585mol = 2788.5 kJ/mol

Entahlpy of Combstion = -2788.5 kJ/mol

dUrxn=  -2788.5 kJ/mol

Formation of glucose:

6C + 3O2 + 6H2 ---> C6H12O6

Since C, O2, and H2 are elemental

Hrxn = Hf

Hf =  -2788.5 kJ/mol

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