6. Sketch and label an energy pathway diagram for this reaction. Your diagram should include: labelling of the X and Y axes ☐ general shape of the energy pathway labeling of the reactant plateau, the product plateau, the activation energy (Ea) and the change in enthalpy (A,H) No specific numerical values are required.

Catalysts affect reaction rates, but do not change either the equilibrium or the spontaneity of a reaction. True or False.

If a reaction is carried out in the presence of a catalyst, will the amount of productpresent at equilibrium increase, decrease, or stay the same?

which statement is true?

Consider the diagraam attached, points A and B represent?     A - intermediate; B - intermediate     A - transition state; B - intermediate     A - intermediate; B - transition state     A - transition state; B - transition stat

Choose the correct equilibrium constant expression corresponding to Kc for the heterogeneous reaction O O Ke о Kc = OKc 2 [H₂01² [0₂][CH4] Kc = Ke 2[H₂O]² [0₂] [CH4] = Ke [02₁] 1 [0₂][CH₁] [H₂O]2[C] [0₂][CH4] CH4(g) + O₂(g) = C(s) + 2H₂O(g)

Problem-Solving Strategies: Understanding Equilibrium ▼ Part A Identify the stress Consider the equilibrium reaction CO(g) + H₂O(g) - CO₂(g) + H₂(g) Determine how each change in the left-hand column will stress the system and in which direction the equilibrium reaction will shift in response. Drag the appropriate labels to their respective targets. ► View Available Hint(s) Increase reactant Decrease reactant Increase product Decrease product Submit Type here to search Reverse reaction rate temporarily increases Forward reaction rate temporarily increases Forward reaction rate temporarily decreases Reverse reaction rate temporarily decreases Change Add CO(g) Remove H₂O(g) Add CO₂(g) Remove H₂(g) I H System stress Group 1 Group 1 Group 1 Group 1 ✔ Ca Equilibrium shift Group 2 Group 2 Group 2 Group 2 Reset Help

Please don't provide handwriting solution

Consider the quilibrium reaction between X and Y, as shown below: X=Y AG The reaction is started with 10 mmol of X; no Y is initially present. After 48 hours, analysis reveals the presence of 10 mmol of X and 0 mmol of Y. Which is the most likely explanation? = −1 - 45 kJ mol X and Y have reached equilibrium concentrations. An enzyme has shifted the equilibrium toward X. Formation of Y is kinetically slow; equilibrium has not been reached by 48 hours. Formation of Y is thermodynamically unfavorable. Two of the above explanations are reasonable.

What is the Law of Mass Action for the following reversible reaction in the gas phase?          H2  +  I2  double arrow  2 HI     [HI]2 / ([I2] X [H2])     2 ([I2] X [H2]) / [HI]      2[HI] / ([I2] + [H2])     ([I2] X [H2])2 / 2 [HI]