# Catalytic activity of transition elements > Why transition elements are ideal catalysts ## Objectives - Read fig 1-3: a catalyst lowers the activation energy; the net energy released stays the same. - Explain why transition elements are ideal catalysts: their 4s and 3d electrons form bonds with reacting molecules on the surface. - Order the book's reason: bonds to the surface, higher concentration, weaker bonds, lower activation energy, faster reaction. - Refute: a catalyst is not used up and gives no extra product or energy. ## 1. Magnetic properties and catalytic activity ## 2. The fizzing powder, again Predict: what does the $\ce{MnO2}$ change? - It lowers the activation energy, the hump ✓ - It reacts with $\ce{H2O2}$ and is used up - It makes the reaction release more energy Fig 1-3: with $\ce{MnO2}$ the hump (the activation energy) is lower, so $\ce{H2O2}$ decomposes faster. The reactants and products are the same, so the net energy released does not change, and the $\ce{MnO2}$ is still there at the end. [ar-EG] شكل 1-3: مع الـ $\ce{MnO2}$ الحدبة (الـ activation energy) بتوطى، فالـ $\ce{H2O2}$ بيتكسّر أسرع. الـ reactants والـ products هما هما، فالـ net energy released مابتتغيرش، والـ $\ce{MnO2}$ بيفضل موجود في الآخر. ## 3. Fig 1-3, step by step Decomposition of hydrogen peroxide (fig 1-3) 1 = no catalyst, 2 = with $\ce{MnO2}$ Effect of MnO₂ as a catalyst in the decomposition of H₂O₂ (fig 1-3) With $\ce{MnO2}$ the hump is lower: a lower activation energy, so a faster reaction. Activation energy is the energy given out by the reaction. ## 4. Lower hump, same drop Look at the two ends of the curves in fig 1-3 below. The catalyst does not move them. Effect of MnO₂ as a catalyst in the decomposition of H₂O₂ (fig 1-3) A catalyst gives more product or more energy. So a catalyst changes only how fast the products form. ## 5. Why transition elements? Book: transition elements are considered ideal catalysts. The reason is in their **4s and 3d electrons**. Book p.15: the catalytic activity of the first transition series ## 6. On the surface stage Molecules arrive. The metal's 4s and 3d electrons form bonds with them. Many molecules are now held on the surface: their concentration there increases. The bonds inside the molecules weaken, so less energy is needed to break them: a lower activation energy. The bonds to the surface do not last. The products leave, and the catalyst is left unchanged. That is why it is not used up. ## 7. Rebuild the chain Put the book's reason in order, from the electrons to the speed. ## 8. New reaction, same rules With the catalyst, how much net energy is released? - 140 kJ: the 80 kJ saved is given out too - 100 kJ, the same as the new activation energy - 60 kJ ✓ Net energy released = the drop from the reactants to the products. The catalyst lowers the activation energy (180 kJ → 100 kJ) but does not move the reactant and product levels, so the net energy released stays 60 kJ. [ar-EG] الـ net energy released = النزلة من الـ reactants للـ products. الـ catalyst وطّى الـ activation energy (من 180 kJ لـ 100 kJ)، لكن ماحرّكش مستوى الـ reactants والـ products، فالـ net energy released فضلت 60 kJ. ## 9. Summary A catalyst lowers the activation energy, so the reaction is faster (fig 1-3). Same products, same net energy released; the catalyst is not used up. Transition elements are ideal catalysts: their 4s and 3d electrons form bonds with reacting molecules on the surface. Decomposition of $\ce{H2O2}$ (fig 1-3) Give reasons: most transition elements have catalytic activity Because of the **4s and 3d electrons**, which form **bonds between the surface atoms and the reacting molecules**. This **increases the concentration** of the molecules on the surface and **weakens their bonds**. So the **activation energy decreases** and the reaction speeds up. Back to the hook: can you explain how $\ce{MnO2}$ speeds up $\ce{H2O2}$ and is still there at the end? Chapter: Transition Elements (https://yaani.ai/ar/learn/chemistry/chem-s3-en.ch1) Source: https://yaani.ai/ar/lesson/chem-s3-en.lesson-te.catalytic-activity