Answer to Question #96220 in Molecular Physics | Thermodynamics for Kaleesh

Question

a) Determine the enthalpy changes, ΔH for the reaction below, given the following reactions and subsequent ΔH values. Please rewrite the amended chemical reaction equation again.

N2H4 (l) + CH4O (l) CH2O (g) + N2 (g) + 3H2 (g)


2NH3 (g) N2H4 (l) + H2 (g) ΔH = 22.5kJ

2NH3 (g) N2 (g) + 3H2 (g) ΔH = 57.5 kJ

CH2O (g) + H2 (g) CH4O (l) ΔH = 81.2 kJ

b) Given that the enthalpy of vaporization for water as below:

H2O (l) H2O (g) ΔH vap = 44.0 kJ mol-1

Calculate enthalpy ΔH for each of the following processes:

i) Evaporating 3.00 moles of water

ii) Evaporating 3.00 grams of water

i) Condensing 20.0 grams of water
(3 Marks)

c) Use the enthalpy of formation data to calculate the enthalpy of the reaction below.

2C2H6 (g) + 7O2 6H2O (g) + 4CO2 (g)

Answer 1

a) -46.2 kJ

b)

(i) 132 kJ/mol

(ii) 7.3 kJ/mol

(iii) -48.9 kJ/mol

c)

$$\Delta H^0 = 4\Delta H(CO_2)+7\Delta H(H_2O) - (2\Delta H(C_2H_6) + 7\Delta H(O_2)) = 4\Delta H(CO_2)+7\Delta H(H_2O) - 2\Delta H(C_2H_6)$$

Because $\Delta H(O_2) = 0 kJ/mol$