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Question: For a reaction \(\dfrac{{dX}}{{dt}} = K{\left[ {{H^ + }} \right]^n}\). If the pH of the reaction med...

For a reaction dXdt=K[H+]n\dfrac{{dX}}{{dt}} = K{\left[ {{H^ + }} \right]^n}. If the pH of the reaction medium changes from two to one, the rate becomes 100{\text{100}} times of the value at pH=2pH = 2. The order of reaction is:
A) 1
B) 2
C) 0
D) 3

Explanation

Solution

The value of pH is the antilog of the H+{H^ + } ions. One can calculate the value of hydrogen ions for pH at 1 and pH at 2 and compare them with the rate equation. The rate of reaction increases 100{\text{100}} times and this can be taken in the equation for rate ratio and calculate the order of the reaction.

Complete step by step answer:

  1. First of all before leading to the solution we will understand what has been given in the question. The first thing is that there have been two pH values given. As per the formula of pH below,
    pH=−log⁡[H+]pH = - \log \left[ {{H^ + }} \right]
    log⁡[H+]=−pH\log \left[ {{H^ + }} \right] = - pH
    That means,
    H+=10−pH{H^ + } = {10^{ - pH}}
    So, the hydrogen ion concentration at pH=1pH = 1,
    H+=10−1{H^ + } = {10^{ - 1}}
    H+=0⋅1{H^ + } = 0 \cdot 1
    Now, for the pH=2pH = 2 hydrogen ion concentration will be,
    H+=10−2{H^ + } = {10^{ - 2}}
    H+=0⋅01{H^ + } = 0 \cdot 01
  2. Now that the rate changes 100{\text{100}} times when pH goes from 2 to 1 it means the concentration of hydrogen ions also changes from 0⋅010 \cdot 01 to 0⋅10 \cdot 1
  3. Now for the pH=1pH = 1 hydrogen ion concentration be 0⋅10 \cdot 1 and the rate will be,
    dxdt=k(0⋅1)n\dfrac{{dx}}{{dt}} = k{\left( {0 \cdot 1} \right)^n}
    Now let's consider the above equation as equation 1.
  4. Now for the pH=2pH = 2 hydrogen ion concentration will be 0⋅010 \cdot 01 and the rate will be,
    dxdt=k(0⋅01)n\dfrac{{dx}}{{dt}} = k{\left( {0 \cdot 01} \right)^n}
    Now let's consider this equation as equation 2.
  5. Now as we need to find out the order of reaction for the pH which changes from 1 to 2 we can calculate that from the below formula by dividing equation 1 by 2,
    100=(0⋅1n0⋅01n)100 = \left( {\dfrac{{0 \cdot {1^n}}}{{0 \cdot {{01}^n}}}} \right)
    We did not take the value of dxdt\dfrac{{dx}}{{dt}} as it gets canceled out due to common in both equations,
    Now let calculate the above equation we get,
    100=10n100 = {10^n}
    Now the value of ‘n’ will be,
    n=2n = 2
    Therefore, the order of reaction will be the second order.
    Hence, option B is the correct choice.

Note: In the above reaction, the pH changes from 1 to 2 which has the order of reaction as 2 which means the reaction rate is dependent on the two reactants and making a change in one of them will alter the rate of reaction.