For the reaction `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` Calculate `Delta_(A)S^(@)`( in `/mol-K)` at 300K

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

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g) , DeltaH = -93.6 KJ mol^(-1) the formation of NH_(3) is expected to increase at :

For the reaction, N_(2)(g) +3H_(2)(g) rarr 2NH_(3)(g) DeltaH =- 95.0 kJ and DeltaS = - 19000 J K^(-1) Calculate the temperature in centigrade at which it will attain equilibrium.

Formation of ammonia is shown by the reaction, N_(2(g))+3H_(2(g)) rarr 2NH_(3(g)), Delta_(f)H^(@)=-"91.8 kJ mol"^(-1) What will be the enthalpy of reaction for the decomposition of NH_(3) according to the reaction ? 2NH_(3(g)) rarr N_(2(g))+3H_(2(g)), Delta_(r)H^(@)=

Assertion (A) : For the reaction N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g) unit of K_(c) =L^(2) mol^(-2) Reason (R ) : For the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) equilibrium constant K_(c)=([NH_(3)]^(2))/([N_(2)]xx[H_(2)]^(3))

For the reaction N_(2)(g) + 3H_(2)(g) rarr 2NH_(2)(g) Which of the following is correct?

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 ?

One "mole" of N_(2) is mixed with three moles of H_(2) in a 4L vessel. If 0.25% N_(2) is coverted into NH_(3) by the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) , calculate K_(c) . Also report K_(c) for 1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)