The diagram in Figure P9.6 is based on the standard heats of formation (Delta H^
ID: 513528 • Letter: T
Question
The diagram in Figure P9.6 is based on the standard heats of formation (Delta H^degree_f) values for four compounds made from the elements listed on the " zero" line of the vertical axis. a. Why are the elements all on the same horizontal line? b. Why is C_2H_2(g) sometimes called an " endothermic" compound? c. Based on these Delta H^degree_F values, predict which of these two reactions has the more negative enthalpy change: 2 C_2H_2(g) + 3 O_2(g) rightarrow 4 CO(g) + 2 H_2O(g) or 2 C_2H_2(g) + 5 O_2(g) rightarrow 4 CO_2(g) + 2 H_2O(g)Explanation / Answer
c) enthalpy of formation of CO2 is -393.5 kJ/mol and enthalpy of formation of CO is -110.5 kJ/mol. Number of moles of C2H2 and moles oh H2O are same both the reactions. Also moles of CO2 and moles of CO are equal.
Hence second reaction which produces CO2 has more negative enthalpy change.
Note: diagram is not given to answer (a) and (b).