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How does the rate of the reaction between hydrogen gas and iodine vapor change with a change in pressure?

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The rate of the reaction between hydrogen gas  H2  and iodine vapor  I2  can be affected by a change in pressure. The reaction can be represented by the following balanced equation:H2 g  + I2 g   2HI g This reaction is a reversible reaction, and according to Le Chatelier's principle, if a system at equilibrium is subjected to a change in pressure, the system will adjust itself to counteract the change and restore equilibrium.If the pressure is increased, the reaction will shift towards the side with fewer moles of gas to counteract the increase in pressure. In this case, the forward reaction will be favored, as there are 2 moles of gas on the right side  2HI  and 2 moles of gas on the left side  H2 + I2 . The rate of the forward reaction will increase, leading to a higher concentration of hydrogen iodide  HI  being formed.If the pressure is decreased, the reaction will shift towards the side with more moles of gas to counteract the decrease in pressure. In this case, the reverse reaction will be favored, as there are 2 moles of gas on the right side  2HI  and 2 moles of gas on the left side  H2 + I2 . The rate of the reverse reaction will increase, leading to a higher concentration of hydrogen gas  H2  and iodine vapor  I2 .In summary, an increase in pressure will favor the forward reaction and increase the rate of HI formation, while a decrease in pressure will favor the reverse reaction and increase the rate of H2 and I2 formation.
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