Explain your answer or show your work: 1. Given the following reaction: 2CO (g)

Question

Explain your answer or show your work:

1. Given the following reaction:

            2CO (g) + O2 (g)--> 2CO2                     H°= -566.0 kJ

Calculate the enthalpy change for the reaction:

            CO (g) + ½ O2 (g)-->CO2

a. -566.8 kJ

b. -283.0 kJ

c. 283.0 kJ

d. 566.0 kJ

e. -1132 kJ

2.Given the following reaction:

            2CO (g) + O2 (g)--->2CO2                     H°= -566.0 kJ

Calculate the enthalpy change for the reaction:

            2CO2 (g)--->2CO (g) + O2 (g)

a. -566.8 kJ

b. -283.0 kJ

c. 283.0 kJ

d. 566.0 kJ

e. -1132 kJ

3.Using the following equations:

C3H8 (g) + 5O2 (g)--->3CO2 (g) + 4H2O (l)                  =-2219.9 kJ

CO (g) + ½ O2 (g)---> CO2 (g)                                      =-283.0 kJ

To calculate the enthalpy change for the reaction:

C3H8 (g) + O2 (g)---> 3CO (g) + 4H2O (l)

a. -1936.9 kJ

b. -1370.9 kJ

c. 1370.9 kJ

d. -3068.9 kJ

e. 3068.9 kJ

Explanation / Answer

1. b. -283.0 kJ (When hal mole reactant will reduce H° to half )

2. d. 566.0 kJ (When we change reactant side H° sign will change)

3. b. -1370.9 kJ

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