This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
Hey K𝔥𝔞𝔫𝔬𝔲𝔪ִֶָ, good to see you again. To calculate the enthalpy change of the reaction ( H_rxn), we use the formula: H_rxn = n H_f^(products) - m H_f^(reactants) where n and m are the stoichiometric coefficients for the products and reactants, respectively. The given reaction is: C_3H_8(g) + 5O_2(g) → 3CO_2(g) + 4H_2O(l) Given standard enthalpies of formation ( H_f^): • H_f^(C_3H_8(g)) = -104 kJ/mol • H_f^(H_2O(l)) = -286 kJ/mol • H_f^(CO_2(g)) = -394 kJ/mol • H_f^(O_2(g)) = 0 kJ/mol (since it is an element in its standard state) Step 1: Calculate the sum of the enthalpies of formation for the products. n H_f^(products) = [3 × H_f^(CO_2(g))] + [4 × H_f^(H_2O(l))] = [3 × (-394 kJ/mol)] + [4 × (-286 kJ/mol)] = -1182 kJ - 1144 kJ = -2326 kJ Step 2: Calculate the sum of the enthalpies of formation for the reactants. m H_f^(reactants) = [1 × H_f^(C_3H_8(g))] + [5 × H_f^(O_2(g))] = [1 × (-104 kJ/mol)] + [5 × (0 kJ/mol)] = -104 kJ + 0 kJ = -104 kJ Step 3: Calculate H_rxn. H_rxn = n H_f^(products) - m H_f^(reactants) = (-2326 kJ) - (-104 kJ) = -2326 kJ + 104 kJ = -2222 kJ/mol The calculated value matches option B. The final answer is B. -2222 kJ/mol. Send me the next one 📸