Determine the change in enthalpy ( H) from Diagram A.

Welcome back Nomvelo — been a while! Let's pick up where you left off. 5.1.1 Step 1: Determine the change in enthalpy ( H) from Diagram A. The potential energy of the reactants is 208.2 kJ. The potential energy of the products is 86.5 kJ. H = E_products - E_reactants H = 86.5 kJ - 208.2 kJ H = -121.7 kJ Step 2: Classify the reaction based on H. Since the change in enthalpy ( H) is negative, the reaction releases energy. The reaction is EXOTHERMIC. Reason: The potential energy of the products (86.5 kJ) is lower than the potential energy of the reactants (208.2 kJ), indicating that energy is released during the reaction. 5.1.2 The shaded area to the right of line P in Diagram B represents the number of particles with kinetic energy equal to or greater than the activation energy. These are the particles that possess sufficient energy to overcome the activation barrier and react. 5.1.3 Step 1: Identify the values from Diagram A for the uncatalyzed reaction. The potential energy of the reactants is 208.2 kJ. The potential energy of the activated complex (transition state) for the uncatalyzed reaction is 240.8 kJ. Step 2: Calculate the activation energy for the uncatalyzed reaction. The letter X in Diagram B represents the activation energy of the uncatalyzed reaction, as it is the higher activation energy. E_a,X = E_activated\ complex - E_reactants E_a,X = 240.8 kJ - 208.2 kJ E_a,X = 32.6 kJ The numerical value represented by the letter X is 32.6 kJ. Send me the next one 📸