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Be sure to answer all parts. For the reaction4A(g) + 3B(g) → 2C(g) The following data were obtained at constant temperature: Experiment Initial [A] (mol/L) Initial [B] (mol/L) Initial Rate(mol/L·min) 1 0.100 0.100 5 2 0.300 0.100 45 3 0.100 0.200 10 4 0.300 0.200 90 Pick the best expression for the rate law for the reaction:

Respuesta :

Answer:

Rate = k [A]2 [B]

Explanation:

4A(g) + 3B(g) → 2C(g)  

Initial [A] (mol/L) Initial [B] (mol/L) Initial Rate(mol/L·min)

1 0.100 0.100 5

2 0.300 0.100 45

3 0.100 0.200 10

4 0.300 0.200 90

Upon observing experiments 2 and 4, along with 1 and 3;

Doubling the concentration of B doubles the rate of the reaction. This means the reaction is first order in respect to B.

Upon observing experiments 1 and 2, along with 3 and 4;

Tripling the concentration of A increases the reaction rate by a factor of 9. This means the reaction is second order with respect to A.

rate law for the reaction:

Rate = k [A]2 [B]

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