Determine the rate law equation for the following reaction, given the experimental data shown below A2+2B=2AB rate=k(B)n(A2)m provided the data: trial run 1 (A2) the A is squared 2 (B) Reaction Rate:
1 1.0M 1.0M 4.0M/s
2 1.0M 2.0M 4.0M/s
3 2.0M 3.0M 8.0 M/s

In this case, given the information on the table, we notice that in the first and second run we left the concentration of A constant whereas the concentration of B was doubled from 1.0M to 2.0M, even do, we notice that the concentration of B does not affect the rate as it remains constant in 4M/s.

Nevertheless, by doubling the concentration of A, we double the rate, that is from 1.0 M to 2.0 M, we are able to rise the rate from 4.0 M/s to 8.0 M/s, for that reason the rate law is zeroth-order with respect to B and first-order with respect to A, that in formula is:

[tex]rate=k[A_2][/tex]

Best regards.

Determine the rate law equation for the following reaction, given the experimental data shown below A2+2B=2AB rate=k(B)n(A2)m provided the data: trial run 1 (A2) the A is squared 2 (B) Reaction Rate: 1 1.0M 1.0M 4.0M/s 2 1.0M 2.0M 4.0M/s 3 2.0M 3.0M 8.0 M/s

Respuesta :

Otras preguntas

Determine the rate law equation for the following reaction, given the experimental data shown below A2+2B=2AB rate=k(B)n(A2)m provided the data: trial run 1 (A2) the A is squared 2 (B) Reaction Rate:
1 1.0M 1.0M 4.0M/s
2 1.0M 2.0M 4.0M/s
3 2.0M 3.0M 8.0 M/s