JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 13)
The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 $$\times$$ 10$$-$$3s$$-$$1 and activation energy is 209 kJ mol$$-$$1. Its rate constant (in s$$-$$1) at 600 K is x $$\times$$ 10$$-$$6. The value of x is ___________. (Nearest integer)
[Given R = 8.31 J K$$-$$ mol$$-$$1; log 6.36 $$\times$$ 10$$-$$3 = $$-$$2.19, 10$$-$$4.79 = 1.62 $$\times$$ 10$$-$$5]
[Given R = 8.31 J K$$-$$ mol$$-$$1; log 6.36 $$\times$$ 10$$-$$3 = $$-$$2.19, 10$$-$$4.79 = 1.62 $$\times$$ 10$$-$$5]
Answer
16
Explanation
K700 = 6.36 $$\times$$ 10$$-$$3s$$-$$1;
K600 = x $$\times$$ 10$$-$$6s$$-$$1
Ea = 209 kJ/mol
Applying;
$$\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$$
$$\log \left( {{{{K_{700}}} \over {{K_{600}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {700}} - {1 \over {600}}} \right)$$
$$\log \left( {{{6.36 \times {{10}^{ - 3}}} \over {{K_{600}}}}} \right) = {{ + 209 \times 1000} \over {2.303 \times 8.31}}\left( {{{100} \over {700 \times 600}}} \right)$$
log(6.36 $$\times$$ 10$$-$$3) $$-$$ logK600 = 2.6
$$\Rightarrow$$ logK600 = $$-$$2.19 $$-$$ 2.6 = $$-$$4.79
$$\Rightarrow$$ K600 = 10$$-$$4.79 = 1.62 $$\times$$ 10$$-$$5
= 1.62 $$\times$$ 10$$-$$6
= x $$\times$$ 10$$-$$6
$$\Rightarrow$$ x = 16
K600 = x $$\times$$ 10$$-$$6s$$-$$1
Ea = 209 kJ/mol
Applying;
$$\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$$
$$\log \left( {{{{K_{700}}} \over {{K_{600}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {700}} - {1 \over {600}}} \right)$$
$$\log \left( {{{6.36 \times {{10}^{ - 3}}} \over {{K_{600}}}}} \right) = {{ + 209 \times 1000} \over {2.303 \times 8.31}}\left( {{{100} \over {700 \times 600}}} \right)$$
log(6.36 $$\times$$ 10$$-$$3) $$-$$ logK600 = 2.6
$$\Rightarrow$$ logK600 = $$-$$2.19 $$-$$ 2.6 = $$-$$4.79
$$\Rightarrow$$ K600 = 10$$-$$4.79 = 1.62 $$\times$$ 10$$-$$5
= 1.62 $$\times$$ 10$$-$$6
= x $$\times$$ 10$$-$$6
$$\Rightarrow$$ x = 16
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