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  • IMAT 2016 Chemistry Q45

  • Ari Horesh

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    June 19, 2021 at 6:26 pm
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    A mixture of methanol and ethanoic acid is left until equilibrium is reached. The equation for this
    reaction is given below.
    The amount of CH3OH in this mixture at equilibrium can be increased by:
    1 adding more water to the mixture
    2 raising the temperature of the mixture
    3 adding sodium hydroxide to the mixture
    4 adding a catalyst to the mixture
    2 and 4 only
    1 and 2 only
    1 and 3 only
    1, 2 and 3 only
    4 only

    • This discussion was modified 3 months ago by  Ari Horesh.
  • Paris Pitsinikos

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    June 19, 2021 at 6:51 pm
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    The reaction is exothermic as demonstrated by the negative enthalpy. The amount of methanol will increase if we add water since the equilibrium will shift towards the side where water is consumed. Raising the temperature will result in the equilibrium shifting towards the endothermic, in order to lower the temperature. This is also going to increase the amount of methanol since the reaction from right to the left is endothermic. Finally, adding sodium hydroxide will result in the neutralisation of the ethanoic acid, thereby ending the forward reaction. Also, taking into account that NaOH will increase the pH, then the equilibrium will shift to decrease it by making more ethanoic acid. Along with the ethanoic acid, the reaction will result in more methanol being made. Hence, 1,2 and 3 are correct so ‘D’ is the correct answer.

    Bobby Sundstrom voted up
  • Ben Naz

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    August 5, 2021 at 12:46 pm
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    <div>
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    This is only for the option 4.

    Adding a catalyst makes absolutely no difference to the position of equilibrium, and Le Châtelier’s principle does not apply. This is because a catalyst speeds up the forward and back reaction to the same extent and adding a catalyst does not affect the relative rates of the two reactions, it cannot affect the position of equilibrium.

    Bobby Sundstrom voted up
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