For a reaction,A_{(g)} to A_{(l)}; Delta H = | vedclass

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(D) The reaction is $A_{(g)} 	o A_{(l)}$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$

Vedclass · Accepted Answer

$|\Delta H| > |\Delta U|$

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(D) The reaction is $A_{(g)} 	o A_{(l)}$.The relationship between enthalpy change and internal energy change is $\Delta H = \Delta U + \Delta n_g RT$.Given $\Delta H = -3RT$.Here,$\Delta n_g = n_{g, 	ext{products}} - n_{g, 	ext{reactants}} = 0 - 1 = -1$.Substituting the values: $\Delta H = \Delta U - RT$.$-3RT = \Delta U - RT$.$\Delta U = -3RT + RT = -2RT$.Comparing the magnitudes: $|\Delta H| = |-3RT| = 3RT$ and $|\Delta U| = |-2RT| = 2RT$.Therefore,$|\Delta H| > |\Delta U|$.

For a reaction,$A_{(g)} \to A_{(l)}$; $\Delta H = -3RT$. The correct statement for the reaction is

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