Use the standard enthalpies of formation to answer the questions on this page. K

Question

Use the standard enthalpies of formation to answer the questions on this page. K_2O_2(s): Delta H degree_f = -496 kJ/mol CO(g): Delta H degree_f =-111 kJ/mol CO_2(g): Delta H degree_f = -394 kJ/mol Write the reactions of formation for the three compounds above. RXN (1) - Formation of K_2O_2 RXN (2) - Formation of CO RXN (3) - Formation of CO_2 What combination of Reactions 1-3 gives the following "reaction X"? K_2O_2(s) + 2 CO(g) rightarrow 2 K(s) + 2 CO_2 RXN X = Calculate Delta H degree_rxn for reaction X. What would be the value of Delta H degree_rxn for: K_2O(s) + CO_2(g) rightarrow K_2O_2(s) + CO(g). given that delta H degree_rxn for K_2O(s) + CO(g) rightarrow 2 K(s) + CO_2(g) is +80 kJ/mol?

Explanation / Answer

Q23

reaction of formation --> must contain only elemental standard states

K2O2 --> O, K

K(s) + O2(g) --> K2O2(g)

balance:

2K(s) + O2(g) --> K2O2(g)

b)

CO --> C and O

C(s) and O2(g) --> CO

balance

C(s) + 1/2O2(g) --> CO(g)

Q3.

CO2 --> C and O

C(s) + O2(g)

balance

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

B)

combination that favours:

K2O2 + 2CO --> 2K + 2CO2

Invert (RXN1) so K2O2 is in the left

Invert (RXN2) and multuply by 2, so 2CO is in the left

Multiply (RXN3) by 2to have 2CO2 in the right

c)

HRxn = -1*(-496) + 2*(-111) + 2(-394) = -514 kJ

d.

find value if we change

K2O + CO = 2K + CO2

invert it...

so

HRxn2 = -HRxn1

HRxn = - 80 kJ/mol

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