4.20 If 3.365 g of ethanol C_2H_5OH(l) is burned completely in a bomb calorimete

Question

4.20

If 3.365 g of ethanol C_2H_5OH(l) is burned completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. a. Calculate Delta H degree_ for ethanol at 298.15 K. b. Calculate Delta H degree_ of ethanol at 298.15 K. From the following data, calculate Delta H degree_ for the reaction CH_3COOH(g) rightarrow 2 H_2O(g) + 2 CO_2(g): Delta H degree _g (kJ mol^-1) CH_3COOH(l) + 2 O_2(g) rightarrow 2 H_2O(l) + 2 CO_2(g) -871.5 H_3O(l) rightarrow H_2O(g) 40.656 CH_3COOH(l) rightarrow CH_3COOH(g) 24.4 Values for Delta H degree_g for the first two reactions are at 298.15 K, and for the third reaction at 391.4 K. Substance C_rho m/R CH_3COOH(g) 14.9 O_2(g) 3.53 CO_2(g) 4.48 H_2O(g) 0.053 H_2O(g) 4.038 A 0.1429

Explanation / Answer

a)

Hcombustion = Qcombustion / moles

Q = 99.472 kJ

n = mass/MW = 3.365/46 = 0.0731521 mol of ethanol

HRcomb = (99.472/0.0731521) = 1359.79 kJ/mol

b)

Hf for ehtanol given:

C2H5OH(l) + O2 (g)= CO2(g) + H2O(g)

balance:

C2H5OH(l) + 3O2 (g)= 2CO2(g) + 3H2O(g)

HRxn = Hproducts - Hreactants

1359.79 = (2*-393.5 + 3*-241.818) - (C2H5OH + 3*0)

C2H5OH = 1359.79- (2*-393.5 + 3*-241.818)

C2H5OH = 2,872.24 kJ/mol

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