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: $CH_4$, $O_2$; Products: $CO_2$, $H_2O$.

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 $O_2$, $\Delta H_f = 0$ (element in standard - state). $\Delta H_{rxn}=[\Delta H_f(CO_2)+2\Delta H_f(H_2O)]-\Delta H_f(CH_4)$ $=[- 393.5+2\times(-241.82)]-(-74.6)$ $=(-393.5 - 483.64)+74.6$ $=-877.14 + 74.6=-802.54\ kJ/mol$.

Step3: Calculate $\Delta H_{rxn}$ for 2 - mol of $CH_4$

Multiply $\Delta H_{rxn}$ for 1 - mol by 2. $\Delta H=-802.54\times2=-1605.08\approx - 1605.1\ kJ$.

Answer:

-1605.1 kJ