Home
Class 12
CHEMISTRY
For urea NH(2) CONH(2) (S) , Delta(f) H(...

For urea `NH_(2) CONH_(2) (S) , Delta_(f) H_(298)^(@) = -333.5 kJ`/mol. Find `Delta_(f) U_(298)^(@)` of urea .

Text Solution

Verified by Experts

–324.8 kJ/mol
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THERMOCHEMISTRY

    MOTION|Exercise Exercise - 4 Level-I|15 Videos
  • THERMOCHEMISTRY

    MOTION|Exercise Exercise - 4 Level-II|8 Videos
  • THERMOCHEMISTRY

    MOTION|Exercise Exercise - 2 (Level-II)|29 Videos
  • SURFACE CHEMISTRY

    MOTION|Exercise Exercise - 3 (Level-II)|12 Videos
  • THERMODYNAMICS

    MOTION|Exercise EXERCISE - 4 (LEVEL - II)|19 Videos

Similar Questions

Explore conceptually related problems

Calculate Delta_(r) H^(@) for Fe_(2) O_(3) (s) + 3 CO(g) to 2 Fe (s) + 3 CO_(2) (g) Given : Delta_(f) H^(@) (Fe_(2) O_(3) , s) = -822.2 kJ/mol Delta_(f) H^(@) (CO , g) = -110.5 kJ//mol , Delta_(f) H^(@) (CO_(2) , g) = -393.5 kJ/mol

Following data is given for the reaction : CaCO_(3) (s) rarr CaO(s) + CO_(2) (g) Delta_(f) H^( Θ) [CaO(s) ] = - 635.1 kJ "mol" ^(-1) Delta_(f) H^(Θ) [ CO_(2) (g) ] = - 393.5 kJ "mol" ^(-1) Delta_(f) H^(Θ)[CaCO_(3)(s) ] = - 1206.9 kJ "mol" ^(-1) Predict the effect of temperature on the equilibrium constant of the above reaction.

Knowledge Check

  • The value of log_10 K for a reaction A to B is Given: Delta_(r) H_(298 K)^(@) = -54.07 kJ "mol"^(-1) , Delta_(r) S_(298 K)^(@) = 10 JK^(-1) "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) 2.303 xx 8.314 xx 298 = 5705

    A
    5
    B
    10
    C
    95
    D
    100
  • The value of log_(10) K for a reaction A iff B is (Given : Delta_(r)H_(298K)^(@)=-54.07 kJ mol^(-1) , Delta_(r)S_(298 K)^(@)=10 JK^(-1) mol^(-1) and R=8.314 JK^(-) mol^(-1), 2.303 xx 8.314xx298=5705 )

    A
    5
    B
    10
    C
    95
    D
    100
  • Similar Questions

    Explore conceptually related problems

    Find the value of log_(10)K for the reaction A Given, Delta_(l)H_(298 K)^(@) = -54.07 kJ "mol"^(-1) Delta_(l)S_(298K)^(@) = 10 JK^(-1) "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) 2.303 xx 8.314 xx 298 = 5705

    The combustion of 1 mol of benzene takes place at 298 K and 1 atm . After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0 kJ of heat is librated. Calculate the standard entalpy of formation, Delta_(f)H^(Θ) of benzene Given: Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1) Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1) .

    The value of log_(10)K for a reaction A hArr B is (Given: Delta_(f)H_(298K)^(Theta) =- 54.07 kJ mol^(-1) , Delta_(r)S_(298K)^(Theta) =10 JK^(=1) mol^(-1) , and R = 8.314 J K^(-1) mol^(-1)

    The enthalpy change for the combustion of N_(2)H_(4)(l) (Hydrazine) is -622.2 kJ mol^(-1) . The products are N_(2)(g) and H_(2)O(l) . If Delta_(f)H^(Theta) for H_(2)O(l) is -285.8 kJ mol^(-1) . The Delta_(f)H^(Theta) for hydrazine is

    Calculate (a) Delta G^(@) and (b) the equilibrium constant for the formation of NO_(2) " from "NO and NO_(2) and O_(2) at 298 K NO (g) +1/2 O_(2) (g) hArr NO_(2) (g) where Delta _(f) G^(@) (NO_(2)) = 52*0 "kJ//mol ", Delta _(f) G^(@) (NO) = 87*0 "kJ//mol," Delta _(f) G^(@) (O_(2)) = 0 "kJ//mol.

    Calculate DeltaG_(reaction) ("kJ"//"mol") for the given reaction at 300 K A_(2)(g)+B_(2)(g)hArr2Ab(g) and at particle pressure of 10^(-2) bar and 10^(-4) Given : Delta H_(f)^(@) AB =180 kJ//mol," "DeltaH_(f)^(@) A_(2)=60 kJ//mol Delta H_(f)^(@) B_(2) = 29.5 kJ//mol," "DeltaS_(f)^(@) AB=210 J//K-mol Delta S_(f)^(@) A_(2) = 190 kJ//mol," "DeltaS_(f)^(@) B_(2)=205 J//K-mol Use : 2.303 Rxx300=5750 "J"//"mole"