Let the rate be given by r = k[A]x[B]y
1) 0.002 = k(0.100)x(0.100)y
2) 0.008 = k(0.100)x(0.200)y
3) 0.018 = k(0.100)x(0.300)y
4) 0.004 = k(0.200)x(0.100)y
5) 0.006 = k(0.300)x(0.100)y
From 1, 2 and 3
0.0080.018=k(0.100)x(0.200)yk(0.100)x(0.300)y
2.25=(1.5)y=>y=2
OR
0.0020.008=k(0.100)x(0.100)yk(0.100)x(0.200)y
4=(2)y=>y=2
From 1, 4 and 5
0.0020.004=k(0.100)x(0.100)yk(0.200)x(0.100)y
2=(2)x=>x=1
OR
0.0020.006=k(0.100)x(0.100)yk(0.300)x(0.100)y
3=(3)x=>x=1
The rate equation for this reaction r = k[A][B]2
From experiment 1 the rate constant
k=0.100 mol dm−3 ∗(0.100 mol dm−3)20.002 mol dm−3 s−1=2 dm6 mol−2 s−1
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