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Question

From the given table for reaction between A and B:
Initial rateInitial rate [A] mol litre1[B] mol litre1at 300 Kat 320 K mol litre1 s1 mol litre1 s1(I) 2.5×1043.0×1055.0×1042.0×103(II)5.0×1046.0×1054.0×103(III) 1.0×1036.0×1051.6×102

Rate constant at 300 K is x×108 mol2 litre2 s1. Find the value of 15x.

A
4
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B
2.66
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C
4.0
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D
4.00
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E
2.67
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Solution

Let the order with respect to A is x and order with respect to B is y.
Then, rate=k[A]x[B]y
5.0×104=k[2.5×104]x[3.0×105]y ..(i)4.0×103=k[5.0×104]x[6.0×105]y ..(ii)1.6×102=k[1.0×103]x[6.0×105]y ..(iii)
Deviding equation (ii) by (iii), we get
0.25=[5.0×101]x
x=2
Deviding equation (i) by (ii) we get,
0.125=(0.5)x(0.5)y
0.125=(0.5)x+Y
x+y=3
Y=1

The rate equation for the reaction is rate=k[A]2[B]

Rate constant (k1) at 300 K,
k1=Rate[A]2[B]=5.0×104 mol litre1 s1(2.5×104 mol litre1)2(3.0×105 mol litre1)=22.5×3×108 mol2 litre2 s1
x=22.5×3
Hence, 15x=4

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