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  1. The position of equilibrium will move in the direction of: the exothermic reaction if the temperature is decreased. the endothermic reaction if the temperature is increased. It may help if you only learn one of these ideas (exothermic or endothermic) because the other is just the other way round.

  2. The reaction is exothermic (from the ΔH provided above). If temperature is increased, the value of K at the higher temperature will be smaller. At the new equilibrium position, the concentration of CH OH will be lower, and the concentrations of H and CO will be higher than in the initial state.

  3. Jun 23, 2023 · Solution. According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts toward the products side. Exercise 13.4.2.

  4. Nov 13, 2022 · The Le Chatelier principle states that the net reaction will be in a direction that tends to reduce the effect of the added H 2. This can occur if some of the H 2 is consumed by reacting with I 2 to form more HI; in other words, a net reaction occurs in the reverse direction.

  5. This phenomenon is summarized by Le Châtelier’s principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium. Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation.

  6. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible.

  7. Aug 14, 2020 · In fact, no matter what the initial concentrations of \(\ce{NO2}\) and \(\ce{N2O4}\) are, at equilibrium the quantity \([\ce{NO2}]^2/[\ce{N2O4}]\) will always be \(6.53 \pm 0.03 \times 10^{-3}\) at 25°C, which corresponds to the ratio of the rate constants for the forward and reverse reactions.