Determine the rate law and the value of k for the following reaction using the data provided.

2n2o5=4no2+o2

[n2o5]                        initial rate

0.093                      4.84 x 10^-4

0.084                      4.37 X 10^-4

0.224                     1.16 x 10^-3

Asked by lekhakarthikeyan | 26th Aug, 2018, 08:00: PM

Expert Answer:

Given:

2 straight N subscript 2 straight O subscript 5 subscript open parentheses straight g close parentheses end subscript end subscript space rightwards arrow space 4 NO subscript 2 subscript open parentheses straight g close parentheses end subscript end subscript space plus space straight O subscript 2 subscript open parentheses straight g close parentheses end subscript end subscript space

 

[N2O5]                      Initial rate

0.093                      4.84 x 10-4   --------- (1)

0.184                      9.37 X 10-4   ---------(2)

0.224                      1.16 x 10-3  ---------(3)

 

From equation (1) and (2), it is observed that when the concentration of [N2O5] doubles the rate is also doubled.

Also, from equation (1) and (3) it is observed that when the concentration of [N2O5] triple the rate is also tripled.

Therefore, 

Rate =k[N2O5]

straight k space equals fraction numerator space 4.84 space straight x space 10 to the power of negative 4 end exponent over denominator 0.093 end fraction

straight k space equals space 5.2 cross times 10 to the power of negative 3 end exponent space straight s to the power of negative 1 end exponent

therefore Rate space law space is space

Rate space equals space 5.2 cross times 10 to the power of negative 3 end exponent space straight s to the power of negative 1 end exponent space open square brackets straight N subscript 2 straight O subscript 5 close square brackets

k = 5.2×10-3 s-1 

Rate law:

Rate =5.2×10-3 s-1 [N2O5]

Answered by Varsha | 27th Aug, 2018, 04:38: PM