Question

Given these reactions,

X(s) + 1/2 O₂(g) -----> XO(s) Delta H = -699.3 kJ/mol

XCO₃(s) -------> XO(s) + CO₂(g) Delta H = +328.3 kJ/mol

what is Delta H_rxn for this reaction?

X(s) + 1/2 O₂(g) + CO₂(g) -------> XCO₃(s) Delta H = kJ/mol

Answer 1

Solution:

X(s) + 1/2 O₂(g) -----> XO(s) Delta H = -699.3 kJ/mol           ….(1)

XCO₃(s) -------> XO(s) + CO₂(g) Delta H = +328.3 kJ/mol         ….(2)

what is Delta H_rxn for this reaction?

X(s) + 1/2 O₂(g) + CO₂(g) -------> XCO₃(s) Delta H = kJ/mol

We have to find delta H of given reaction.

Lets reverse reaction 2

XO(s) + CO₂(g) ------->   XCO₃(s) Delta H = - 328.3 kJ/mol        ….(3)

(When we reverse reaction the sign of delta H changes)

Lets add reaction 3 and 1

X(s) + 1/2 O₂(g) -----> XO(s) Delta H = -699.3 kJ/mol          

XO(s) + CO₂(g) ------->   XCO₃(s) Delta H = - 328.3 kJ/mol  

--------------------------------------------      

X (s) +1/2 O2 (g) + CO2 (g) --- > XCO3 (s)       Delta H = -699.3 kJ/mol +    - 328.3 kJ/mol  

                                                                                              =     -1027.6   kJ/mol

(when we add reactions then their delta H should be added. )

Answer : -1027.6 kJ/mol