From the following data: C (graphite) + O2 (g) --> CO2(g) H rxn = -303.5 kJ/mol

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                     

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.