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Assertion: For Adsorption DeltaG,DeltaH,...

Assertion: For Adsorption `DeltaG,DeltaH,DeltaS` all have -ve values.
Reason: Adsorption is a exothermic process in which randomness decreases due to force of attraction between adsorbent and adsorbate.

A

Statement I is true, Statement II is also true and Statement II is the correct explanation of Statement I.

B

Statement I is true, Statement II is also true and Statement II is not the correct explanation of Statement I.

C

Statement I is true, Statement II is false

D

Statement I is false, Statement II is true.

Text Solution

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The correct Answer is:
To solve the assertion-reason question regarding adsorption, we will analyze both the assertion and the reason step by step. ### Step 1: Understanding the Assertion The assertion states that for adsorption, the values of ΔG (Gibbs free energy change), ΔH (enthalpy change), and ΔS (entropy change) are all negative. 1. **ΔG (Gibbs Free Energy Change)**: For a process to be spontaneous, ΔG must be negative. Adsorption is a spontaneous process, which means ΔG is indeed negative. 2. **ΔH (Enthalpy Change)**: Adsorption is an exothermic process, meaning it releases heat. In exothermic reactions, ΔH is negative. 3. **ΔS (Entropy Change)**: Entropy (ΔS) is a measure of randomness or disorder in a system. During adsorption, gas molecules lose their freedom of movement when they adhere to a surface, leading to a decrease in randomness. Therefore, ΔS is negative. ### Conclusion for Assertion Since all three values (ΔG, ΔH, ΔS) are negative, the assertion is correct. ### Step 2: Understanding the Reason The reason states that adsorption is an exothermic process where randomness decreases due to the attractive forces between the adsorbent and adsorbate. 1. **Exothermic Process**: As stated, adsorption releases energy, confirming that ΔH is negative. 2. **Decrease in Randomness**: When gas molecules are adsorbed onto a solid surface, their movement is restricted, leading to a decrease in randomness. This confirms that ΔS is negative. ### Conclusion for Reason The reason provided correctly explains why the assertion is true. Therefore, both the assertion and the reason are correct, and the reason is a valid explanation for the assertion. ### Final Answer Both the assertion and reason are correct, and the reason is the correct explanation for the assertion.
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Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. A reaction has a value of DeltaH =-40 Kcal at 400 k cal mol^(-1) . The reaction is spontaneous, below this temperature , it is not . The values fo DeltaG and DeltaS at 400 k are respectively

Dependence of Spontaneity on Temperature: For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. DeltaG_(P.T) lt 0. DeltaG_(P.T) =0 implies the equilibrium condition and DeltaG_(P.T) gt 0 corresponds to non- spontaneity. Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : " "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1) The magnitude of DeltaH does not change much with the change in temperature but the entropy factor TDeltaS change appreciably . Thus, spontaneity of a process depends very much on temperature. For endothermic process, both DeltaH and DeltaS are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor TDeltaS will be small and may be less than DeltaH, DeltaG will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor TDeltaS Increases appreciably and when it exceeds DeltaH, DeltaG would become negative and the process would be spontaneous . For an expthermic process, both DeltaH and DeltaS would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when T DeltaS gt DeltaH, DeltaG will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor , TDeltaS decreases rapidly and when TDeltaS lt DeltaH, DeltaG becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature. For the reaction at 298 K ,2A + B rarr C DeltaH =100 kcal and DeltaS=0.050 kcal K^(-1) . If DeltaH and DeltaS are assumed to be constant over the temperature range, above what temperature will the reaction become spontaneous?