JEE MAIN - Chemistry (2020 - 6th September Morning Slot - No. 7)
The variation of equilibrium constant with
temperature is given below :
The values of $$\Delta $$Ho, $$\Delta $$Go at
T1 and $$\Delta $$Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
| Temperature | Equilibrium Constant |
|---|---|
| T1 = 25oC | K1 = 10 |
| T2 = 100oC | K2 = 100 |
The values of $$\Delta $$Ho, $$\Delta $$Go at
T1 and $$\Delta $$Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
28.4, –5.71 and –14.29
0.64, –7.14 and –5.71
28.4, –7.14 and –5.71
0.64, –5.71 and –14.29
Explanation
ln $$\left[ {{{{k_2}} \over {{k_1}}}} \right]$$ = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}$$
$$ \Rightarrow $$ ln(10) = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {298}} - {1 \over {373}}} \right\}$$
$$ \Rightarrow $$ $${\Delta H^\circ }$$ = 28.37 kJ/mol
$$\Delta $$Go = –RT ln K
T1 = 25oC K1 = 10
$$\Delta $$Go at T1 = –8.314 × 298 × 2.303 × log 10
= –5.71 kJ/mol
$$\Delta $$Go at T2 = –8.314 × 373 × 2.303 × log(100)
= –14.29 kJ/mol
$$ \Rightarrow $$ ln(10) = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {298}} - {1 \over {373}}} \right\}$$
$$ \Rightarrow $$ $${\Delta H^\circ }$$ = 28.37 kJ/mol
$$\Delta $$Go = –RT ln K
T1 = 25oC K1 = 10
$$\Delta $$Go at T1 = –8.314 × 298 × 2.303 × log 10
= –5.71 kJ/mol
$$\Delta $$Go at T2 = –8.314 × 373 × 2.303 × log(100)
= –14.29 kJ/mol
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