For the equilibrium, `N_2O_4 iff 2NO_2, DeltaG "_(N_2O_4)^0 = 100 kJ mol^-1` and `DeltaG "_(NO_2)^0 = 50kJ mol^-1` at 298 K. When 5 mol / lit of each is taken then find the value of `DeltaG` for the reaction at.298 K

At 298K, for attainment of equilibrium of the reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) , 5 mol of each of the constituents is taken. Due to this, total pressure of the mixture turns 20 atm. If DeltaG_(f)^(0)(N_(2)O_(4))=100kJ*mol^(-1) and DeltaG_(f)^(0)(NO_(2))=50K(J*mol^(-1)) then- (1) Give the value of DeltaG of the reaction? (2) In which direction will the reaction proceed more to attain equilibrium?

Given C (graphite) + O_(2) (g) to CO_(2) (g) Delta ""_(r) H^(0) = -393.5 k J * mol^(-1) H_(2) (g) + (1)/(2) O_(2) (g) to H_(2) O (l) , Delta ""_(r) H^(0) = -285.8 kJ * mol^(-1) CO_(2) (g) + 2 H_(2) O(l) to CH_(4) + 2 O_(2) (g) , Delta ""_(r)H^(0) = + 890 . 3 kJ * mol^(-1) Based on the above thermochemical equations , the value of Delta ""_(r) H^(0) at 298 K for the reation C(graphite) + 2 H_(2) (g) to CH_(4) (g) will be -

For the reaction X_2O_4(l) to 2XO_2(g) , DeltaU = 2.1 kcal, DeltaS = 20 cal K^-1 at 300k Hence, DeltaG is

For the reaction at 298 k,2A +B rarr C, Delta H= 400 KJ mol^(-1) and Delta S=0.2 KJ k^(-1) mol^(-1). At what temperature will the reaction become spontaneous ?

Calculate the temperature at which DeltaG = -5.2 kJ mol^-1, DeltaH = 145.6 kJ mol^-1 and DeltaS = 216 J K^-1 mol^-1 for a chemical reaction

For the reaction, 2A+B to C,DeltaH=400kJ*mol^(-1) & DeltaS=0.2kJ*K^(-1)*mol^(-1) at 298K. At what temperature will the reaction become spontaneous considering DeltaH,DeltaS to be constant over the temperature range?

In the reaction N_(2(g)) + O_(2(g)) to 2NO_((g)), DeltaH^@ reaction is + 179.9 kJmol^(-1) and DeltaS_("reaction")^@ = 66.09 JK^(-1)mol^(-1) . Calculate DeltaG^@ reaction at 300K.

Calculate the DeltaG^0 for the reaction N_2(g) + O_2(g) * 2NO(g) Given DeltaH^0 =180.5 K.J , and DeltaS^0 = 15 J at 25^@C .