For the reaction, `2A rarr A_2` following data is obtained at 298K. `triangleU = -20KJ`, `triangleS= -30J` then find `triangle G`

For the non-stoichiometric reaction 2A +B rarr C+D , The following data were obtained in three separate experiments, all at 298 K. The rate law for the formation of C is:

For the reaction, 2A (s) + 2B (g)to C (g) + D(s) at 298K Delta U ^(@) =-10.5KJ and Delta S ^(@)=-44.10 JK^(-1) Calculate Delta G ^(@) for the reaction (in Joules).

If for the reaction [2A(I) rarr 4B(g)] delta U and deltaS respectively are 1.2 kcal and 10 cal K^-1 at 300 K then delta G for the reaction is

For the reaction, 2NOCl(g) hArr 2NO(g)+Cl_(2)(g) Calculate the standard equilibrium constant at 298 K . Given that the value of DeltaH^(ɵ) and DeltaS^(ɵ) of the reaction at 298 K are 77.2 kJ mol^(-1) and 122 J K^(-1) mol^(-1) .

For a demerization reaction , 2A (g) to A_2 (g) at 298 K, Delta U(Theta) = - 20 kJ"mol"^(-1), Delta S^(Theta) = - 30 JK^(-1) "mol"^(-1) , then the Delta G^(Theta) will be _________J.

For the reaction X_(2)O_(4(l)) rarr 2XO_(2(g)), Delta U= 2.1 "kcal", Delta S= 20 "cal " K^(-1) at 300K, Hence Delta G is

Consider the reaction: 2NO(g)+O_(2)(g) rarr 2NO_(2)(g) Calculated the standard Gibbs energy change at 298K and predict whether the reaction is spontaneous or not. Delta_(f)G^(Theta) (NO) = 86.69 kJ mol^(-1), Delta_(f)G^(Theta) (NO_(2)) = 51.84 kJ mol^(-1) .

Consider the reaction: 2NO(g)+O_(2)(g) rarr 2NO_(2)(g) Calculated the standard Gibbs energy change at 298K and predict whther the rection is spontaneous or not. Delta_(f)G^(Theta) (NO) = 86.69 kJ mol^(-1), Delta_(f)G^(Theta) (NO_(2)) = 51.84 kJ mol^(-1) .

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.

For the reaction at 25^(@)C, C_(2)O_(4)(l) rarr 2XO_(2)(g) Delta H =2.1 kcal and DeltaS=20 cal K^(-1) . The reaction would be :