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Find out the internal energy change for the reaction `A (l) rarr A (g)` at 373 K . Heat of vaporisation is 40.66 kJ `//` mol and `R = 8.3 J mol^(-1) K^(-1)`

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To find the internal energy change (ΔU) for the reaction \( A (l) \rightarrow A (g) \) at 373 K, we can use the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) given by the equation: \[ \Delta H = \Delta U + \Delta N_g RT \] Where: - \( \Delta H \) is the heat of vaporization (given as 40.66 kJ/mol). ...
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Calculate the standard free energy change for the reaction: H_(2)(g) +I_(2)(g) rarr 2HI(g), DeltaH^(Theta) = 51.9 kJ mol^(-1) Given: S^(Theta) (H_(2)) = 130.6 J K^(-1) mol^(-1) , S^(Theta) (I_(2)) = 116.7 J K^(-1) mol^(-1) and S^(Theta) (HI) =- 206.8 J K^(-1) mol^(-1) .

90 g of water spilled out from a vessel in the room on the floor. Assuming that water vapour behaving as an ideal gas, calculate the internal energy change when the spilled water undergoes complete evaporation at 100^(@)C . (Given the molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol^(-1) ).

Knowledge Check

  • Find the entropy change for vaporisation of water to steam at 100^@C in JK^(-1)mol^(-1) if heat of vaporisation is 40.8 kJ mol^(-1) .

    A
    `109.38`
    B
    `100.38`
    C
    `110.38`
    D
    `120.38`

  • Evaluate the entropy change in the vaporisation of 36 gm water at 373 K. (Delta H_("vap")=40.63 kJ mol^(-1))

    A
    `210 JK^(-1)`
    B
    `218 JK^(-1)`
    C
    `216 JK^(-1)`
    D
    `118 JK^(-1)`

  • Assuming that water vapour is an ideal gas, the internal energy change (DeltaU) when 1 mole of water is vaporised at 1 bar pressure and 100^@ C , (given : molar enthalpy of vaporization of water "41 kJ mol"^(-1) at 1 bar and 373 k and R = 8.3 "J mol"^(-1) K^(-1) ) will be

    A
    4.100 `"kJ mol"^(-1)`
    B
    3.7904 `"kJ mol"^(-1)`
    C
    37.904 `"kJ mol"^(-1)`
    D
    41.00 `"kJ mol"^(-1)`

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