The Enthalpy of combustion of C6H5‌COOH(s) at 25∘c and 1.0 atm pressure is −2546‌kJ∕mol. What is the Internal energy change for this reaction?
To find the internal energy change, ∆U, we use the relationship between the change in enthalpy, ∆H, and ∆U in a chemical reaction. The relationship is given by: ∆H=∆U+∆ngasRT Here: ∆H is the enthalpy change. ∆U is the internal energy change. ∆n‌gas ‌ is the change in the number of moles of gas during the reaction. R is the universal gas constant, which is approximately 8.314Jmol−1K−1 or 0.008314‌kJ‌mol−1‌K−1. T is the temperature in Kelvin. Given the temperature is 25∘C, we convert it to Kelvin as 25+273.15=298.15K The standard combustion reaction of benzoic acid is: C6H5‌COOH(s)+7.5O2(g)⟶7CO2(g)+3H2O(l) From the reaction, the change in the number of moles of gas, ∆n‌gas ‌, can be calculated as: ∆n‌gas ‌= moles of gaseous products - moles of gaseous reactants So, ∆n‌gas ‌=7−7.5=−0.5 So, ∆n‌gas ‌=7−7.5=−0.5 Now, substituting these values into the equation, we get: ‌∆H=∆U+∆ngasRT ‌−2546‌kJ=∆U+(−0.5)(0.008314‌kJ‌mol−1‌K−1)(298.15K) ‌−2546‌kJ=∆U−1.238‌kJ ‌∆U=−2546‌kJ+1.238‌kJ ‌∆U=−2544.762‌kJ Rounding to the nearest decimal, we get:
∆U≈−2544.8‌kJ Therefore, the correct answer is: Option A: -2544.8 kJ