Calculate the internal energy at 298 K for the formation of one mole of ammonia if the enthalpy change at constant pressure is - 42.0 `kJ mol^(-1)` (G

Question

Calculate the internal energy at 298 K for the formation of one mole of ammonia if the enthalpy change at constant pressure is - 42.0 `kJ mol^(-1)` (G

Calculate the internal energy at 298 K for the formation of one mole of ammonia if the enthalpy change at constant pressure is - 42.0 `kJ mol^(-1)` (Given `: R = 8.314 JK^(-1) mol^(-1)) `

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

Standard enthalpy of vapourisation `DeltaH^(@)` forwater is `40.66 KJ mol^(-1)`.The internal energy of vapourisation of water for its 2 mol will be :-

2 Answers 4 Views

Calculate the change in entropy for the process : \( H _{2} O _{\left(\text {liq, } 20^{\circ} C , 1 atm \right)} \) to \( H _{2} O _{\left.\text {(gas, } 250^{\circ} C , 1 atm \right)} \) You are given, heat capacity of water \( ( \) liq \( )=75.6 JK ^{-1} \cdot mol ^{-1} \) Heat capacity of water \( ( \) vap \( )=36.2 JK ^{-1} mol ^{-1} \) Heat of vaporization of \( H _{2} O \) at \( 100^{\circ} C \) and \( 1 atm 40.85 JK mol ^{-1} \)

2 Answers 4 Views

When one mole of monoatomic ideal gas at T temperature undergoes adiabatic change under a constant external pressure of 1 atm, change in volume is fro

2 Answers 5 Views

Heat of formation of `H_(2)O` is -188kJ/mol and `H_(2)O_(2)` is -286 kJ/mol. The enthalpy change for the reaction, `2H_(2)O_(2) to 2H_(2)O+O_(2)`

2 Answers 5 Views

The standard enthalpy of formation of `NH_(3)` is `-46.0kJmol^(-1)`. If the enthalpy of formation of `H_(2)` from its atoms is `-436kJmol^(-1)`, the a

2 Answers 4 Views

Calculate the enthalpy of vaporisation of `I_(2)` if the sublimation energy and enthalpy of fusion of `I_(2)` is 57.3 kJ `mol^(-1)` and 15.5 kJ `mol^(

2 Answers 4 Views

Identify the equation in which change in enthalpy is equal to change in internal energy

2 Answers 4 Views

At STP, a container has 1 mole of Ar, 2 mole of `CO_(2)`, 3 moles of `O_(2)` and 4 moles of `N_(2)`. Without changing the total pressure if one mole o

2 Answers 5 Views

The `DeltaH` for vaporization of a liquid is 20kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the li

2 Answers 4 Views

The molar enthalpy change for `H_(2)O(l)hArrH_(2)O(g)` at 373 K and 1 atm ius 41 kJ/mol. Assume ideal behaviour, the internal energy change for vapori

2 Answers 4 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`Delta U = - 39. 523 KJ`

5 views Answered 2023-02-05 15:29