Question

The decomposition of N2O5 in the gas phase was studied at constant temperature.

2 N 2O5(g) → 4 NO2(g) + O2(g)

The following results were collected:

[N 2O5] Time (s)

0.1000 0

0.0707 50

0.0500 100

0.0250 200

0.0125 300

0.00625 400


(a) Determine the rate law and calculate the value of k.   



(b) Determine the concentration of N2O5(g) at 250 s. Does your calculated answer make sense? Justify your answer.



(c) Determine the concentration of N2O5(g) at 600 s. Does your calculated answer make sense? Justify your answer.



(d) At what time is the concentration of N2O5(g) equal to 0.00150 M ? Explain how you know your answer is correct.








Ans: rate = k[N2O5] ; k = −slope of graph of ln [N2O5] vs. time = 6.93 × 10−3 s−1; 0.0177 M; 0.00156

Answer 1

(a) The plots of [N2O5] vs t, ln[N2O5] vs t and 1/[N2O5] vs t are drawn below, of which only ln[N2O5] vs t plot is straight line plot with R² =1, rest are not. This happens only in case of first order reaction. Thus order of the reaction is first order.

The integrated rate law of first order is ln[N2O5] = ln[N2O5]₀ -kt, so slope of the plot of ln[N2O5] vs t would give the value of rate constant,k. Thus k= -0.00693 s^-1 and rate law is rate= k[N2O5].

(b) Given [N2O5]₀(initial concentration) = 0.1 M, so from the integrated rate law mentioned above, putting t= 250 s and k=0.00693 s^-1 we get, ln[N2O5] = ln(0.1) -0.00693s^-1x250s = -4.0276, whih gives [N2O5] = e^(-4.0276) = 0.0177 M, which is infact almost in the middle of the concentrations given at t=200 and 300 s. Thus the value found is justified.

(c) [N2O5] for t= 600 is found just like above.

ln[N2O5] = ln(0.1) -0.00693s^-1x600s = -6.4606, which on solving gives [N2O5] = 0.00156 M, which is almost 1/4 times the concentration at 400 s, and from the trend around 300 and 400 s we see it to be halving every 100 s, so in 200 s past 400 s it should become almost 1/4 times of what it was at 400 s, that is 0.00625/4 = 0.00156 M, which ,matches the value found by us above. Hence the answer is justified.

(d) ln(0.00150) = ln(0.1) - 0.00693s^-1x(t), which gives t= 606 s. Our answer is correct because in part (c) we proved that concentration near 600 s is 0.00156 M.

Comment in case of any doubt.