7. What is meant by thermodynamie standard state? . Calculate the heat absorbed

Question

7. What is meant by thermodynamie standard state? . Calculate the heat absorbed by 500.0 g of nickel when its temperature is -warned from 22.4 C to 58.4 C (at constant pressure). The specific heat of nickel is 0.44 J/(g." C) 9. Calculate the heat of reaction of ethyl alcohol and oxygen CllsOll(l) + 302(g) 2C0:(g) + 31120(1) Standard enthalpies of formation, AH'r (kcal/mol) ChisOH(l)--65.9 kcal/nnol; CO2(g)--941 kealinol; keal/mol 110(1)--68.1 Hint: AH' rnHr (products) -YAH: (reactants) = Using the data below; calculate the enthalpy change for the following reaction: 3NO2(g) + H2O(l) = 2HNO3(aq) + NO(g) H'r: NO2(g) = 33. 10 kJ/mol; H2O(1)--285.8 kJ/mol; INOdaq)--207.4 kJ/mol; NO(g) = 90.29 kJ

Explanation / Answer

Ans. #7. The thermodynamic standard states include-

                        Temperature = 0.00C = 273.15 K

                        Pressure = 105 Pa

                        Physical state: The natural state of the specified substance. For example, it I “solid for water” and “gas for oxygen”

#8. The amount of heat gained by is given -

            q = m s dT                          

Where,

q = heat

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in above reaction-

            q = 500.0 g x (0.44 J g-10C-1) x (58.4 – 22.4)0C

            Hence, q = 7920.0 J

Therefore, heat absorbed by Ni sample = 7920.0 J

#9. dHrxn = Sum of standard enthalpies of formation of (products – reactants)

            Or, dHrxn = [(2 x dHf0 CO2,g) + (3 x dHf0 H2O,l)] - [(dHf0 C2H5OH,l) + (3 x dHf0 O2,g)]

            Or, dHrxn = [ (2 x (-94.1) + 3 x (-68.1)] kcal/mol - [ (-65.9) + (3 x 0)] kcal/mol

            Or, dHrxn = (-392.5 kcal/mol) – (-65.9 kcal/mol)

            Hence, dHrxn = - 326.6 kcal/mol

#10. dHrxn = Sum of standard enthalpies of formation of (products – reactants)

            Or, dHrxn = [(2 x dHf0 HNO3,aq) + (dHf0 NO,g)] - [(3 x dHf0 NO2,g) + (dHf0 H2O,l)]

            Or, dHrxn = [(2 x (-207.4) + 90.29] kJ/mol - [3 x 33.10 + (-285.8)] kJ/mol

            Or, dHrxn = (-324.51 kJ/mol) – (-186.5 kJ/mol)

            Hence, dHrxn = -138.01 kJ/mol

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