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.\nc(s) + 2o₂(g) → co₂(g)\nwhat is the enthalpy change of the reaction?\nuse △hrxn = ∑(△hf,products) - ∑(△hf,reactants).\n-393.5 kj\n-196.8 kj\n196.8 kj\n393.5 kj

carbon reacts with oxygen to produce carbon dioxide (co₂(g), △hf = -393.5 kj/mol) according to the equation below.\nc(s) + 2o₂(g) → co₂(g)\nwhat is the enthalpy change of the reaction?\nuse △hrxn = ∑(△hf,products) - ∑(△hf,reactants).\n-393.5 kj\n-196.8 kj\n196.8 kj\n393.5 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 an element in its standard state is 0 kJ/mol. So, $\Delta H_{f,C(s)} = 0$ kJ/mol and $\Delta H_{f,O_2(g)}=0$ kJ/mol, and $\Delta H_{f,CO_2(g)}=- 393.5$ kJ/mol.

Step2: Apply the enthalpy - change formula

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

Answer:

-393.5 kJ