JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 17)
2NO(g) + Cl2(g) $$\rightleftharpoons $$ 2NOCl(s)
This reaction was studied at $$-$$10$$^\circ$$ and the following data was obtained
$${[NO]_0}$$ and $${[C{l_2}]_0}$$ are the initial concentrations and r0 is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).
This reaction was studied at $$-$$10$$^\circ$$ and the following data was obtained
| Run | $${[NO]_0}$$ | $${[C{l_2}]_0}$$ | $${r_0}$$ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.18 |
| 2 | 0.10 | 0.20 | 0.35 |
| 3 | 0.20 | 0.20 | 1.40 |
$${[NO]_0}$$ and $${[C{l_2}]_0}$$ are the initial concentrations and r0 is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).
Answer
3
Explanation
We know,
$$Rate = K{\left[ A \right]^x}{\left[ B \right]^y}$$
Here from,
Exp 1 : $$0.18 = K{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}$$ ..... (1)
Exp. 2 : $$0.35 = K{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}$$ .... (2)
Exp. 3 : $$1.40 = K{\left[ {0.2} \right]^x}{\left[ {0.2} \right]^y}$$ .... (3)
(2) $$\div$$ (3)
$${{0.35} \over {1.40}} = {{K \times {{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}} \over {K \times {{\left[ {0.2} \right]}^x}{{\left[ {0.2} \right]}^y}}}$$
$$ \Rightarrow {1 \over 4} = {\left( {{1 \over 2}} \right)^x}$$
$$ \Rightarrow x = 2$$
(1) $$\div$$ (2)
$$\left( {{1 \over 2}} \right) = {\left( {{1 \over 2}} \right)^y}$$
$$ \Rightarrow y = 1$$
$$ \therefore $$ x + y = 3
$$Rate = K{\left[ A \right]^x}{\left[ B \right]^y}$$
Here from,
Exp 1 : $$0.18 = K{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}$$ ..... (1)
Exp. 2 : $$0.35 = K{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}$$ .... (2)
Exp. 3 : $$1.40 = K{\left[ {0.2} \right]^x}{\left[ {0.2} \right]^y}$$ .... (3)
(2) $$\div$$ (3)
$${{0.35} \over {1.40}} = {{K \times {{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}} \over {K \times {{\left[ {0.2} \right]}^x}{{\left[ {0.2} \right]}^y}}}$$
$$ \Rightarrow {1 \over 4} = {\left( {{1 \over 2}} \right)^x}$$
$$ \Rightarrow x = 2$$
(1) $$\div$$ (2)
$$\left( {{1 \over 2}} \right) = {\left( {{1 \over 2}} \right)^y}$$
$$ \Rightarrow y = 1$$
$$ \therefore $$ x + y = 3
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