JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 10)
For a reaction consider the plot of $$\ell $$n k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is –
_12th_January_Evening_Slot_en_10_1.png)
10$$-$$4 s$$-$$1
4 $$ \times $$ 10$$-$$4 s$$-$$1
10$$-$$6 s$$-$$1
2 $$ \times $$ 10$$-$$4 s$$-$$1
Explanation
From Arhenius equation,
K = Ae$$^{ - {{Ea} \over {RT}}}$$
taking log both sides,
lnk = lnA $$-$$ $${{{Ea} \over {RT}}}$$
So in lnK vs $${1 \over T}$$ graph
slope = $$-$$ $${{{Ea} \over R}}$$
Rate constant at 400 K is = 10$$-$$5 s$$-$$1
Let rate constant at 500 K is = K
$$ \therefore $$ ln(10$$-$$5) = ln A $$-$$ $${{Ea} \over {R\left( {400} \right)}}$$ . . . (1)
ln K = ln A $$-$$ $${{Ea} \over {R\left( {500} \right)}}$$ . . . .(2)
Subtracting (2) from (1) we get,
ln $${K \over {{{10}^{ - 5}}}}$$ = $${{{Ea} \over R}}$$ $$\left[ {{1 \over {400}} - {1 \over {500}}} \right]$$
$$ \Rightarrow $$ 2.303 log $${K \over {{{10}^{ - 5}}}}$$ = 4606 $$\left[ {{1 \over {400}} - {1 \over {500}}} \right]$$
$$ \Rightarrow $$ K = 10$$-$$4 s$$-$$1
K = Ae$$^{ - {{Ea} \over {RT}}}$$
taking log both sides,
lnk = lnA $$-$$ $${{{Ea} \over {RT}}}$$
So in lnK vs $${1 \over T}$$ graph
slope = $$-$$ $${{{Ea} \over R}}$$
Rate constant at 400 K is = 10$$-$$5 s$$-$$1
Let rate constant at 500 K is = K
$$ \therefore $$ ln(10$$-$$5) = ln A $$-$$ $${{Ea} \over {R\left( {400} \right)}}$$ . . . (1)
ln K = ln A $$-$$ $${{Ea} \over {R\left( {500} \right)}}$$ . . . .(2)
Subtracting (2) from (1) we get,
ln $${K \over {{{10}^{ - 5}}}}$$ = $${{{Ea} \over R}}$$ $$\left[ {{1 \over {400}} - {1 \over {500}}} \right]$$
$$ \Rightarrow $$ 2.303 log $${K \over {{{10}^{ - 5}}}}$$ = 4606 $$\left[ {{1 \over {400}} - {1 \over {500}}} \right]$$
$$ \Rightarrow $$ K = 10$$-$$4 s$$-$$1
Comments (0)
