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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

The correct Answer is:
`-87.42kJ `
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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 ?

Knowledge Check

  • The enthalpy change (DeltaH) for the reaction, N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) " is "-92.38 kJ at 298 K. The internal energy change DeltaU at 298 K is

    A
    `-92.38 kJ`
    B
    `-87.42 kJ`
    C
    `-97.34 kJ`
    D
    `-89.9 kJ`

  • 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 mol^(-)`
    B
    `- 87.42 kJ`
    C
    `- 97.34 kJ`
    D
    `- 89.9 kJ`

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

    A
    atm
    B
    `(atm)^(3)`
    C
    `(atm)^(-2)`
    D
    `(atm)^(2)`

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