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

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

A

`-92.38 kJ`

B

`-87.42 kJ`

C

`-97.34 kJ`

D

`-89.9 kJ`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaH=DeltaU+Deltan_(g)RT`
`Deltan_(g)=-2`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • THERMODYNAMICS AND THERMOCHEMISTRY

    ERRORLESS |Exercise ORDINARY THINKING (Objective Questions) Heat of reaction|174 Videos

  • THERMODYNAMICS AND THERMOCHEMISTRY

    ERRORLESS |Exercise ORDINARY THINKING (Objective Questions) Bond energy|15 Videos

  • THERMODYNAMICS AND THERMOCHEMISTRY

    ERRORLESS |Exercise JEE section (JEE (Advanced) 2018) Numeric answer type questions|2 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    ERRORLESS |Exercise JEE Section (Matrix Match type questions)|3 Videos

Similar Questions

Explore conceptually related problems

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 ?

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

Knowledge Check

  • Enthalpy change for the reaction , 4H(g) rarr 2H_(2)(g) is -869.6 kJ . The dissociation energy of H-H bondis

    A
    `+217 .4kJ`
    B
    `-434.8 kJ`
    C
    `-869.6kJ`
    D
    `+ 434.8kJ`

  • Enthalpy change for the reaction, 4H_((g))rarr2H_(2(g)) is - 869.6 kJ. The dissociation energy of H - H bond is

    A
    `+ 217.4` kJ
    B
    `- 434.8` kJ
    C
    `- 869.6` kJ
    D
    `+ 434.8` kJ

    • Similar Questions

    Explore conceptually related problems

    The enthalpy change (DeltaH) for the reaction. N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g) is -92.38 kJ at 298 K. What is DeltaE at 298 K?

    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 internalenergy change ( Delta U ) for the reaction CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l) is - 885kJ mol^(-1) at298 K . What is DeltaH at298 K ?

    What two changes on the equilibrium, N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) , Delta H= -92.4" kJ." can keepitsstate undisturbed ?

    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)

    The standard Gibbs energy change for the reaction N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g) is -33.2 kJ "mol"^(-1) at 298 K. (a) Calculate the equilibrium constant for the above reaction. (b) What would be the equilibrium constant if the reaction is written as 1/2N_(2)(g) + 3/2H_(2)(g) iff NH_(3)(g) (c) What will be the equilibrium constant if the reaction is NH_(3)(g) iff 1/2N_(2)(g) + 3/2 H_(2)(g) .

    Home
    Profile