JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 1)
For the chemical reaction X $$\rightleftharpoons$$ Y, the standard reaction Gibbs energy depends on temperature T (in K) as
$$\Delta $$rGo (in kJ mol–1) = 120 $$ - {3 \over 8}$$ T.
The major component of the reaction mixture at T is :
The major component of the reaction mixture at T is :
Y if T = 300 K
Y if T = 280 K
X if T = 350 K
X if T = 315 K
Explanation
X $$\rightleftharpoons$$ Y
Keq = $${{\left[ Y \right]} \over {\left[ X \right]}}$$
If Keq > 1 $$ \Rightarrow $$ [Y] > [X]
and Keq < 1 $$ \Rightarrow $$ [Y] < [X]
We know, $$\Delta $$Go = -RT ln(Keq)
So when Keq > 1 then $$\Delta $$Go < 0 and Y is major.
And when Keq < 1 then $$\Delta $$Go > 0 and X is major.
Temperature at which Go = 0 is
120 $$ - {3 \over 8}$$ T = 0
$$ \Rightarrow $$ T = 320 K
For T > 320 K
$$\Delta $$Go < 0 and Y is major.
For T < 320 K
$$\Delta $$Go > 0 and X is major.
Keq = $${{\left[ Y \right]} \over {\left[ X \right]}}$$
If Keq > 1 $$ \Rightarrow $$ [Y] > [X]
and Keq < 1 $$ \Rightarrow $$ [Y] < [X]
We know, $$\Delta $$Go = -RT ln(Keq)
So when Keq > 1 then $$\Delta $$Go < 0 and Y is major.
And when Keq < 1 then $$\Delta $$Go > 0 and X is major.
Temperature at which Go = 0 is
120 $$ - {3 \over 8}$$ T = 0
$$ \Rightarrow $$ T = 320 K
For T > 320 K
$$\Delta $$Go < 0 and Y is major.
For T < 320 K
$$\Delta $$Go > 0 and X is major.
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