JEE Main 2020 — Chemistry Physical Chemistry

$${T}_{1}=323K{T}_{2}=373K$$T1​=323KT2​=373K

$${k}_{1}=10{k}_{2}=100$$k1​=10k2​=100

$$\mathrm{log}(\frac{{k}_{2}}{{k}_{1}})=\frac{\Delta H}{2.303R}[\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}]$$log(k1​k2​​)=2.303RΔH​[T1​1​−T2​1​]

$$\mathrm{log}(\frac{100}{10})=\frac{\Delta H}{2.303\times 8.314}[\frac{1}{298}-\frac{1}{373}]$$log(10100​)=2.303×8.314ΔH​[2981​−3731​]

$$\mathrm{log}10=\frac{\Delta H}{2.303\times 8.314}[\frac{75}{298\times 373}]$$log10=2.303×8.314ΔH​[298×37375​]

$$\Delta H=\frac{2.303\times 8.314\times 298\times 373}{75}=28.4\mathrm{KJ}$$ΔH=752.303×8.314×298×373​=28.4KJ

At ${T}_{1}=25^{\circ}C=298K,{K}_{1}=10$T1​=25∘C=298K,K1​=10

$$\Delta G=-2.303{\mathrm{RT}}_{1}{\text{logk }}_{1}$$ΔG=−2.303RT1​logk 1​

$$=-2.303\times 8.314\times 298\times \mathrm{log}(10)$$=−2.303×8.314×298×log(10)

$$=-2.303\times 8.314\times 298\times 1$$=−2.303×8.314×298×1

$$=-5.7\mathrm{KJ}$$=−5.7KJ

At ${T}_{2}=100^{\circ}C=373K{K}_{2}=100$T2​=100∘C=373KK2​=100

$$\Delta G=-2.303{\mathrm{RT}}_{2}{\mathrm{logK}}_{2}$$ΔG=−2.303RT2​logK2​

$$=-2.303\times 8.314\times 373\times \mathrm{log}(10{)}^{2}$$=−2.303×8.314×373×log(10)2

$$=-2.303\times 2\times 8.314\times 373\times 1$$=−2.303×2×8.314×373×1

$$=-14283.7J$$=−14283.7J

$$=-14.29\mathrm{KJ}$$=−14.29KJ