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From the following data: C (graphite) + O2 (g) --> CO2(g) H rxn = -303.5 kJ/mol
From the following data: C (graphite) + O2 (g) --> CO2(g) H rxn = -303.5 kJ/mol
ID:
686438
• Letter:
F
Question
From the following data: C (graphite) + O2 (g) --> CO2(g) H rxn = -303.5 kJ/mol H2 (g) + 1/2 O2 (g) --> H2O(l) H rxn = -285.8 kJ/mol 2 C2H6 (g) + 7 O2 --> 4 CO2 (g) 6 H2O (l) rxn = -3119.6 kJ/mol Calculate the enthalpy change for the reaction 2 C (graphite) + 3 H2 (g) --> C2H6 (g) From the following data: C (graphite) + O2 (g) --> CO2(g) H rxn = -303.5 kJ/mol H2 (g) + 1/2 O2 (g) --> H2O(l) H rxn = -285.8 kJ/mol 2 C2H6 (g) + 7 O2 --> 4 CO2 (g) 6 H2O (l) rxn = -3119.6 kJ/mol Calculate the enthalpy change for the reaction 2 C (graphite) + 3 H2 (g) --> C2H6 (g)
Explanation / Answer
C(graphite) + O2 (g)--> CO2(g) H rxn = -303.5 kJ/mol H2 (g) + 1/2 O2 (g) --> H2O(l) Hrxn = -285.8 kJ/mol 2 C2H6 (g) + 7 O2 --> 4 CO2 (g) 6 H2O(l) rxn = -3119.6 kJ/mol Fromthe given data , 2C(graphite) + 2O2 (g) --> 2CO2(g) H rxn = -(303.5 x2) kJ/mol -------- (1) =- 607 kJ/mol 3H2 (g) + 3/2 O2 (g) --> 3H2O(l) Hrxn = -(285.8 x 3) kJ/mol ------ (2) =-857.4 kJ/mol 2 CO2 (g) + 3 H2O (l) --> C2H6 (g) + (7/2) O2 rxn= (3119.6/2) kJ/mol = 1559.8 kJ/mol by adding equation (1) , (2) and(3) we get 2 C (graphite) + 3 H2 (g) --> C2H6(g) ; Hrxn = -607 kJ/mol + -857.4 kJ/mol + 1559.8 kJ/mol = 95.4 kJ/mol H2 (g) + 1/2 O2 (g) --> H2O(l) Hrxn = -285.8 kJ/mol 2 C2H6 (g) + 7 O2 --> 4 CO2 (g) 6 H2O(l) rxn = -3119.6 kJ/mol Fromthe given data , 2C(graphite) + 2O2 (g) --> 2CO2(g) H rxn = -(303.5 x2) kJ/mol -------- (1) =- 607 kJ/mol 3H2 (g) + 3/2 O2 (g) --> 3H2O(l) Hrxn = -(285.8 x 3) kJ/mol ------ (2) =-857.4 kJ/mol 2 CO2 (g) + 3 H2O (l) --> C2H6 (g) + (7/2) O2 rxn= (3119.6/2) kJ/mol = 1559.8 kJ/mol by adding equation (1) , (2) and(3) we get 2 C (graphite) + 3 H2 (g) --> C2H6(g) ; Hrxn = -607 kJ/mol + -857.4 kJ/mol + 1559.8 kJ/mol = 95.4 kJ/mol 2 C (graphite) + 3 H2 (g) --> C2H6(g) ; Hrxn = -607 kJ/mol + -857.4 kJ/mol + 1559.8 kJ/mol = 95.4 kJ/mol
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