JEE MAIN - Chemistry (2020 - 2nd September Evening Slot - No. 13)
The results given in the below table were
obtained during kinetic studies of the following
reaction
2A + B $$ \to $$ C + D_2nd_September_Evening_Slot_en_13_1.png)
X and Y in the given table are respectively :
2A + B $$ \to $$ C + D
X and Y in the given table are respectively :
0.3, 0.4
0.4, 0.3
0.4, 0.4
0.3, 0.3
Explanation
Rate = k[A]a[B]b
6 × 10–3 = k(0.1)a(0.1)b ...(1)
2.4 × 10–3 = k(0.1)a(0.2)b ...(2)
1.2 × 10–3 = k(0.2)a (0.1)b ...(3)
(3) ÷ (1) $$ \Rightarrow $$ x = 1
(2) ÷ (3) $$ \Rightarrow $$ x = 2
So, order with respect to A = 1
Order with respect to B = 2
6 × 10–3 = k(0.1)1 (0.1)2 from (1)
$$ \Rightarrow $$ k = 6
From experiment IV,
7.2 × 10–2 = 6(x)1(0.2)2 $$ \Rightarrow $$ x = 0.3
From experiment V,
2.88 × 10–2 = 6(0.3)1(y)2 $$ \Rightarrow $$ y = 0.4
6 × 10–3 = k(0.1)a(0.1)b ...(1)
2.4 × 10–3 = k(0.1)a(0.2)b ...(2)
1.2 × 10–3 = k(0.2)a (0.1)b ...(3)
(3) ÷ (1) $$ \Rightarrow $$ x = 1
(2) ÷ (3) $$ \Rightarrow $$ x = 2
So, order with respect to A = 1
Order with respect to B = 2
6 × 10–3 = k(0.1)1 (0.1)2 from (1)
$$ \Rightarrow $$ k = 6
From experiment IV,
7.2 × 10–2 = 6(x)1(0.2)2 $$ \Rightarrow $$ x = 0.3
From experiment V,
2.88 × 10–2 = 6(0.3)1(y)2 $$ \Rightarrow $$ y = 0.4
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