Home
Class 11
CHEMISTRY
For N(2) + 3H(2) rarr 2NH(3), enthalpy a...

For `N_(2) + 3H_(2) rarr 2NH_(3)`, enthalpy and internal energy changes respectively are, `Delta H " & " Delta U` then

A

`Delta H = O`

B

`Delta H Delta U`

C

`Delta H lt Delta U`

D

`Delta H gt Delta U`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

Chemical reactions are invariably associated with the transfer of energy either in the form of hear or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as: q= ms Delta T , s= Specific heat = c Delta T = Heat capacity. Heat of reaction at constant pressure is measured using simple or water calorimeter. Q_(v)= Delta U = Internal energy change, Q_(P) = DeltaH, Q_(P) = Q_(V) + P Delta V and DeltaH = Delta U+ Delta nRT . The amount of energy released during a chemical change depends on the physical state of reactants and products, the condition of pressure, temperature and volume at which the reaction is carried out. The variation of heat of reaction with temperature and pressure is given by Kirchoff's equation: (DeltaH_(2) - DeltaH_(1))/(T_(2)-T_(1))= Delta C_(P) (At constant pressure), (DeltaU_(2) - DeltaU_(1))/(T_(2)-T_(1)) = DeltaC_(V) (At constant volume) The enthalpy change (DeltaH) for the reaction N_(2) (g) + 3H_(2)(g) rarr 2NH_(3)(g) is -92.38kJ at 298 K. The internal energy change DeltaU at 298 K is

N_(2) + 3H_(2) rarr 2NH_(3), Delta H = - 46K . Cals. From the above reaction, heat of formation of ammonia is

For the reaction, N_(2) + 3H_2 rarr 2NH_(3) if E_1and E_2 equivalent masses of NH_3 and N_2 respectively then E_1 -E_2 is

For N_(2) + 3 H_(2) hArr 2 NH_(3) , continous removal of NH_(3) maintains the following condition

For N_(2)+3H_(2)to2NH_(3) rates of disappearance of N_(2) and H_(2) and rate of appearance of NH_(3) respectively, are a,b and c then

(A) : For N_(2) + 3H_(2) hArr 2NH_(3). Delta H =-Q KJ, high pressure yiclds more Ammonia (R) According to Lechatlier's principle, increase of pressure shifts equilibrium in a direction that proceeds in decrease in number of moles.

One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 90K) rarr (4.0 atm, 5.0 L, 245 K) with a change in internal energy, Delta U = 30.0 L atm . The change in enthalpy (Delta H) of the process in L atm is (Give your answer after dividing with 11)

H_2 + Cl_2 rarr 2HCl , DeltaH_1, N_2 + 3H_2 rarr 2NH_3, DeltaH_2 and NH_3 + 3Cl_2 rarr NCl_3 + 3HCI, DeltaH_3 . Then calculate the heat of formation of nitrogen trichloride.