JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 21)
For the reaction A(g) $$ \to $$ B(g) the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of $$\Delta $$rG for the reaction at 300 K and 1 atm in J mol-1 is – xR, where x is _______. (Rounded off to the nearest integer)
[R = 8.31 J mol–1K-1 and ln 10 = 2.3)
[R = 8.31 J mol–1K-1 and ln 10 = 2.3)
Answer
1380
Explanation
For a reaction, A(g) $$\to$$ B(g)
Given, Kp (equilibrium constant) = 100
Temperature = 300 K
Pressure = 1 atm
Formula used, $$\Delta$$G$$^\circ$$ = $$-$$ RT ln Kp .... (i)
Here, $$\Delta$$G$$^\circ$$ = standard Gibb's free energy
R = gas constant = 8.31 J mol$$-$$1 K$$-$$1
Put value in Eq. (i), we get
$$\Delta$$G$$^\circ$$ = $$-$$ R (300) ln 100
$$\Delta$$G$$^\circ$$ = $$-$$ R (300) (2) ln (10)
$$\because$$ ln (10) = 2.3
$$\Delta$$G$$^\circ$$ = $$-$$ R(300) (2) (2.3)
$$\Delta$$G$$^\circ$$ = $$-$$ 1380 R
Hence, $$\Delta$$G$$^\circ$$ = $$-$$ xR
$$ \therefore $$ x = 1380
Given, Kp (equilibrium constant) = 100
Temperature = 300 K
Pressure = 1 atm
Formula used, $$\Delta$$G$$^\circ$$ = $$-$$ RT ln Kp .... (i)
Here, $$\Delta$$G$$^\circ$$ = standard Gibb's free energy
R = gas constant = 8.31 J mol$$-$$1 K$$-$$1
Put value in Eq. (i), we get
$$\Delta$$G$$^\circ$$ = $$-$$ R (300) ln 100
$$\Delta$$G$$^\circ$$ = $$-$$ R (300) (2) ln (10)
$$\because$$ ln (10) = 2.3
$$\Delta$$G$$^\circ$$ = $$-$$ R(300) (2) (2.3)
$$\Delta$$G$$^\circ$$ = $$-$$ 1380 R
Hence, $$\Delta$$G$$^\circ$$ = $$-$$ xR
$$ \therefore $$ x = 1380
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