Answer:
Given reaction is
UO2(NO3)2(aq) -----> UO3(s) + 2NO2(g) + 1/2O2(g)
Zero-order rate law is
[UO2(NO3)2]=-kt+[UO2(NO3)2]0 and the plot of [UO2(NO3)2] vs t gives a straight line with slope=-k.
The first-order rate law is
ln[UO2(NO3)2]=-kt+ln[UO2(NO3)2]0 and the plot of ln[UO2(NO3)2] vs t gives a straight line with slope=-k.
The second-order rate law is
1/[UO2(NO3)2]=kt+1/[UO2(NO3)2]0 and the plot of 1/[UO2(NO3)2] vs t gives a straight line with slope=k.
Where k=rate constant, [UO2(NO3)2]0=Initial concentration at t=0 min
[UO2(NO3)2]=concentration of UO2(NO3)2 at time t.
The calculated values are tabulated below
Time(s) | [UO2(NO3)2] (M) | ln[UO2(NO3)2] | 1/[UO2(NO3)2] |
0 | 1.41E-02 | -4.259455082 | 70.77140835 |
20 | 1.10E-02 | -4.513502997 | 91.24087591 |
60 | 7.58E-03 | -4.882242079 | 131.9261214 |
180 | 3.02E-03 | -5.802498448 | 331.1258278 |
360 | 5.50E-04 | -7.50559228 | 1818.181818 |
[UO2(NO3)2] vs t plot is
ln[UO2(NO3)2] vs t plot is
1/[UO2(NO3)2] vs t plot is
From the above plots, the plot of ln[UO2(NO3)2] vs t gives a straight line with R2=0.997.
Therefore the order of the reaction is first order.
The slope of the line=-k=-0.00882 min-1
Therefore rate constant, k=0.00882 min-1.
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