Question

A scientist measures the standard enthalpy change for the following reaction to be 2752.8 kJ : 6CO2(g) + 6H2O(1) C6H12O6 + 6 O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CH1:05 is kJ/mol

Answer 1

1)

Given:

ΔHo rxn = 2752.8 KJ/mol

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(l)) = -285.83 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Balanced chemical equation is:

6 CO2(g) + 6 H2O(l) ---> C6H12O6 + 6 O2(g)

ΔHo rxn = 1*Hof(C6H12O6) + 6*Hof(O2(g)) - 6*Hof( CO2(g)) - 6*Hof(H2O(l))

2752.8 = 1*Hof(C6H12O6) + 6*(0.0) - 6*(-393.509) - 6*(-285.83)

Hof(C6H12O6) = -1323.234 KJ/mol

Answer: -1323.2 KJ/mol

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