a. Write an equation for the formation of HCl(g) from its elements in its standa

Question

a. Write an equation for the formation of HCl(g) from its elements in its standard states.

b. Find Hf for HCl(g) from Appendix IVB in the textbook.

c. Write an equation for the formation of CO2(g) from its elements in its standard states.

d. Find Hf for CO2(g) from Appendix IVB in the textbook.

e. Write an equation for the formation of Fe2O3(s) from its elements in its standard states.

f. Find Hf for Fe2O3(s) from Appendix IVB in the textbook.

g. Write an equation for the formation of C2H2(g) from its elements in its standard states.

h. Find Hf for C2H2(g) from Appendix IVB in the textbook.

Explanation / Answer

Ans:

A substance's standard enthalpy of formation is the enthalpy change that occurs when one mole of that substance is formed from its constituent elements in their standard state.

Horeaction = Hof(products)Hof(Reactants)

a) 1/2H2(g) + 1/2Cl2(g) -----> HCl(g)

b) Hf for HCl(g) = 92.30 kJ/mol (because Horeaction = 92.30 - [1/2 (0) + 1/2 (0)] = 92.30

c) C(s) (graphite) + O2(g) ------> CO2(g)

d) Hf for CO2(g) = 393.509 kJ/mol ( Horeaction = 393.509 - [1/2 (0) + 1/2 (0)] = 393.509 kJ/molCarbon naturally exists as graphite and diamond. The enthalpy difference between graphite and diamond is too large for both to have a standard enthalpy of formation of zero. To determine which form is zero, the more stable form of carbon is chosen. This is also the form with the lowest enthalpy, so graphite has a standard enthalpy of formation equal to zero.)

e) 2Fe(s) + 3/2O2(g) ------> Fe2O3(s)

f) Hf for Fe2O3(s) = 824.2 kJ/mol ( Horeaction = 842.2 - [1/2 (0) + 1/2 (0)] = 842.2 kJ/mol)

g) 2C(s) + H2(g) ------> C2H2(g)

h) Hf for C2H2(g) = 226.73 kJ/mol (Horeaction = 226.73 - [1/2 (0) + 1/2 (0)] = 226.73 kJ/mol)

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