carbon reacts with oxygen to produce carbon dioxide (co2(g), △hf = -393.5 kj/mol) according to the equation…

carbon reacts with oxygen to produce carbon dioxide (co2(g), △hf = -393.5 kj/mol) according to the equation below. c(s) + 2o2(g)→co2(g) what is the enthalpy change of the reaction? use △hrxn = ∑(△hf,products) - ∑(△hf,reactants). -393.5 kj -196.8 kj 196.8 kj 393.5 kj

carbon reacts with oxygen to produce carbon dioxide (co2(g), △hf = -393.5 kj/mol) according to the equation below. c(s) + 2o2(g)→co2(g) what is the enthalpy change of the reaction? use △hrxn = ∑(△hf,products) - ∑(△hf,reactants). -393.5 kj -196.8 kj 196.8 kj 393.5 kj

Answer

Explanation:

Step1: Identify reactants and products

Reactants are C(s) and O₂(g), product is CO₂(g). The standard - enthalpy of formation of C(s) in its standard state is 0 kJ/mol and of O₂(g) in its standard state is 0 kJ/mol, and $\Delta H_f$ for CO₂(g) is - 393.5 kJ/mol.

Step2: Apply the enthalpy - change formula

$\Delta H_{rxn}=\sum(\Delta H_{f,products})-\sum(\Delta H_{f,reactants})$. Substituting the values: $\Delta H_{rxn}=\Delta H_f(CO_2)-[\Delta H_f(C) + 2\Delta H_f(O_2)]$. Since $\Delta H_f(C)=0$ kJ/mol and $\Delta H_f(O_2) = 0$ kJ/mol, then $\Delta H_{rxn}=-393.5-(0 + 2\times0)=-393.5$ kJ/mol.

Answer:

-393.5 kJ