6C(graphite) + 6H2(g) +3O2(g) >> C6H12O6(s) Delta H1 = -1247 kJ mol^-1
C(graphite) + O2(g) >> CO2(g) Delta H2 = -393.5 kJ mol^-1
H2(g) +1/2O2(g) >> H2O(l) Delta H3 = -286.0 kJ mol^-1
a.Construct an enthalpy change cycle to determine the standard enthalpy change of combustion for glucose.
b.Construct an enthalpy level diagram for the combustion of glucose.
2. Given that the standard enthalpy of formation of liquid methanol, carbon dioxide gas and liquid water are -238.5, -393.5 and -285.8 kJ mol^-1.
a.Determine the standard enthalpy change of combustion of liquid methanol.
3.Given that:
C2H2(g) + 2H2(g) >> C2H6(g) Delta H1 = -311.0 kJ mol^-1
H2(g) + 1/2O2(g) >> H2O(l) Delta H2 = -286.0 kJ mol^-1
C2H6(g) + 7/2O2(g) >> 2CO2(g) +3H2O(l) Delta H3 = -1560 kJ mol^-1
a.Construct an enthalpy change cycle to determine the standard enthalpy change of combustion.
4.100cm^3 of 1.0M sodium hydroxide solution and 100 cm^3 of 1.0M ethanoic acid were mixed in a calorimeter. The maximum temperature rise was found to be 5.3 celcius degree. (given that the specfic heat capacity and the density of the reaction mixture are 4.2 J g^-1 K^-1 and 1.0gcm^-3 respectively, the heat capacity of calorimeter is 90.0JK^-1)
a.Calculate the standard enthalpy change of neutralization of sodium hydroxide solution and ethanoic acid.
b.The theoretical value of the standard enthalpy change of neutralization for a strong acid and a strong alkali is -57.3 kJmol^-1. Account for the discrepancy between the theoretical value and the value obtained in (a).
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更新1:
1a. C6H12O6 + 6O2 >> 6CO2 + 6H2O Enthalpy change of combustion for glucose: 6x(-393.5)+6x(-286)-(-1274) = -2806 Is that right? I know how to write the equation but I don't know how to draw cycle and level diagram, can someone help me?