Question

Consider the following reaction: Ca(s) + 2 H_2O(l) rightarrow Ca(OH)_2(s) + H_2(g) Calculate the heat of reaction based on the following information: 2H_2(g) + O_2(g) rightarrow 2 H_2O(l) DeltaH = -572 kJ/mol CaO(s) + H_2O(l) rightarrow Ca(OH)_2(s) DeltaH = -64 kJ/mol CaCO_3(s) rightarrow CaO(s) + CO_2(g) DeltaH = +178.1 kJ/mol 2 Ca (s) + O_2(g) rightarrow 2 CaO(s) DeltaH = -1270 kJ/mol 13. Acetylene is used in blow torches, and bums according to the following equation: 2 C_2H_2(g) + 5 O_2(g) rightarrow 4 CO_2(g) + 2 H_2O(g) Use the following information to calculate the heat of reaction: DeltaH_o^f (H_2O(g)) = -241.82 kJ/mol DeltaH_o^f (CO_2(g)) = -393.5 kJ/mol DeltaH_o^f (C_2H_2(g)) = 226.77 kJ/mol

Answer 1

Problem 12

1) 2H₂(g) + O₂(g) $\rightarrow$ 2H₂O(l) -----------------   $\Delta$ H = - 572 kJ/mol

Reverse it and divide it by 2 we get

1.a) H₂O(l) $\rightarrow$ H₂(g) + (1/2)O₂(g) ---------------   $\Delta$ H = 286 kJ/mol

Second equation we will use as is

2) CaO(s) + H₂O(l) $\rightarrow$ Ca(OH)2(s) --------------- $\Delta$ H =  - 64 kJ/mol

3) We dont need 3rd equation

4) 2Ca(s) + O₂(g) $\rightarrow$ 2CaO(s) ---------------   $\Delta$ H = - 1270 kJ/mol   

Divide equation 4 by 2, we get

4.a) Ca(s) + (1/2) O₂(g) $\rightarrow$ CaO(s) ---------------   $\Delta$ H = - 635 kJ/mol   

By adding 1.a, 2 and 4.a and its $\Delta$ H values we get the required equation and its $\Delta$ H

Ca(s) + 2H₂O(l) $\rightarrow$ Ca(OH)₂(s) + H₂(g) ---------------   $\Delta$ H = - 413 kJ/mol

Problem 13

Heat of the reaction = Heat of formation of products - heat of formation of reactants

So

Heat of the reaction = 4 * $\Delta$ H^f of CO_2(g) + 2 * $\Delta$ H^f of H₂O(g) - 2 * $\Delta$ H^f of C₂H₅(g)

Heat of the reaction = 4 * (-393.5) + 2 * (-242.82) - 2 * 226.77

Heat of the reaction = - 2511.18 kJ/mol