the enthalpies of formation of the compounds in the combustion of methane, ch4(g)+2o2(g)→co2(g)+2h2o(g), are…

the enthalpies of formation of the compounds in the combustion of methane, ch4(g)+2o2(g)→co2(g)+2h2o(g), are ch4(g): δhf = -74.6 kj/mol; co2(g): δhf = -393.5 kj/mol; and h2o(g): δhf = -241.82 kj/mol. how much heat is released by the combustion of 2 mol of methane? use δhrxn=∑(δhf,products) - ∑(δhf,reactants). -80.3 kj -802.5 kj -1,605.1 kj -6,420.3 kj

the enthalpies of formation of the compounds in the combustion of methane, ch4(g)+2o2(g)→co2(g)+2h2o(g), are ch4(g): δhf = -74.6 kj/mol; co2(g): δhf = -393.5 kj/mol; and h2o(g): δhf = -241.82 kj/mol. how much heat is released by the combustion of 2 mol of methane? use δhrxn=∑(δhf,products) - ∑(δhf,reactants). -80.3 kj -802.5 kj -1,605.1 kj -6,420.3 kj

Answer

Explanation:

Step1: Identify reactants and products

Reactants: $CH_4(g), O_2(g)$; Products: $CO_2(g), H_2O(g)$

Step2: Calculate $\Delta H_{rxn}$ for 1 - mol of $CH_4$

Using $\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})$. For $O_2(g)$, $\Delta H_f = 0$ kJ/mol. $\Delta H_{rxn}=[\Delta H_f(CO_2(g)) + 2\times\Delta H_f(H_2O(g))]-[\Delta H_f(CH_4(g))+2\times\Delta H_f(O_2(g))]$ $=[- 393.5+2\times(-241.82)]-[-74.6 + 2\times0]$ $=(-393.5-483.64)-(-74.6)$ $=-393.5 - 483.64 + 74.6$ $=-802.54$ kJ/mol

Step3: Calculate heat released for 2 - mol of $CH_4$

Multiply $\Delta H_{rxn}$ for 1 - mol by 2. $q = 2\times\Delta H_{rxn}=2\times(-802.54)\approx - 1605.1$ kJ

Answer:

-1,605.1 kJ