Ask Question
4 October, 17:58

Experiment Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L. s 1 0.0500 0.0500 0.0100 6.25 x 10-3 2 0.100 0.0500 0.0100 1.25 x 10-2 3 0.100 0.100 0.0100 2.50 x 10-2 4 0.0500 0.0500 0.0200 1.25 x 10-2What is the rate law for the reaction? a) k[A][B][C]

b) k [A]2[B][C]

+3
Answers (1)
  1. 4 October, 20:47
    0
    k[A][B][C]

    Explanation:

    Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L. s

    1 0.0500 0.0500 0.0100 6.25 x 10-3

    2 0.100 0.0500 0.0100 1.25 x 10-2

    3 0.100 0.100 0.0100 2.50 x 10-2

    4 0.0500 0.0500 0.0200 1.25 x 10-2

    Comparing equations 1 and 2, the reaction is first order with respect to A. This is because the concentration of A doubles, while the concentration of B and C remained constant leading to a double of the rate of reaction.

    Comparing equations 2 and 3, the reaction is first order with respect to B. This is because the concentration of B doubles, while the concentration of A and C remained constant leading to a double of the rate of reaction.

    Comparing equations 1 and 4, the reaction is first order with respect to C. This is because the concentration of C doubles, while the concentration of A and B remained constant leading to a double of the rate of reaction.

    This means our rate law = k[A][B][C];

    That is first order with respect to A, B and C
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Experiment Initial [A], mol/L Initial [B], mol/L Initial [C], mol/L Initial rate, mol/L. s 1 0.0500 0.0500 0.0100 6.25 x 10-3 2 0.100 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers