Home
Class 12
CHEMISTRY
The enthalpy change (Delta H) for the re...

The enthalpy change `(Delta H)` for the reaction
`N_(2) (g)+3H_(2)(g) rarr 2NH_(3)(g)`
is `-92.38 kJ` at `298 K`. What is `Delta U` at `298 K`?

Text Solution

Verified by Experts

`-87.425 kJ`
Promotional Banner

Topper's Solved these Questions

  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise EXERCISE 3|1 Videos
  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise EXERCISE 4|1 Videos
  • THERMODYNAMICS AND THERMOCHEMISTRY

    FIITJEE|Exercise EXERCISE 1|1 Videos
  • TEST PAPERS

    FIITJEE|Exercise CHEMISTRY|747 Videos
  • TRANSITION ELEMENTS & COORDINATION COMPOUNDS

    FIITJEE|Exercise MATCHIG LIST TYPE QUESTIONS|1 Videos

Similar Questions

Explore conceptually related problems

The enthalpy change (DeltaH) for the reaction, NH_(2(g))+3H_(2(g)) rarr 2NH_(3g) is -92.38kJ at 298K What is DeltaU at 298K ?

The enthalpy change (Delta H) for the reaction N_(2) (g) + 3 H_(2) (g) rarr 2 NH_(3) (g) is - 92.38 kJ at 298 K . The internal energy change Delta U at 298 K is

For the reaction, N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , the units of K_(p) are …………

The value of Delta H for the reaction 2 N_(2)(g) + O_(2) (g) rarr 2N_(2) O (g) at 298 K is 164 kJ . Calculate Delta U for the reaction. Strategy : In this process, 3 mol of gas change to 2 mol of gas at constant temperature and pressure. Assuming ideal gas behavior, we can use. First Delta_(n_(g)) and the obtain a value of Delta U by converting the value of Delta H from 164 kJ to 164000 J and expressing R in units of J mol^(-1) K^(-1)

For the reaction 2NH_(3)(g) hArr N_(2)(g) +3H_(2)(g) the units of K_(p) will be

Calculate the standard free energy change for the reaction, N_(2(g))+3H_(2(g))to2NH_(3(g)) at 298 K . Given, DeltaH^@ =-92.4kJ and DeltaS^@ = -198.3 JK^(-1) . Also, comment on the result.