The value of DeltaG^(@) for a reaction is 7.97 KJ//"mole" . Its equilibrium constant K_(c) at 298K is 4 xx 10^(-x) What is x?
For a chemical reaction, the standard Gibbs energy change, DeltaG^(@) is - 7.64xx10^(4) J "mol"^(-1) . What is the value of equilibrium constant (K) ?
Calculate a) DeltaG^(@) and b) the equilibrium constant for the formation of NO_(2) from NO and O_(2) at 298K NO(g)+1//2O_(2)(g)hArrNO_(2)(g) where Delta_(f)G^(oplus)(NO_(2))=52.0kJ//mol Delta_(f)G^(oplus)(NO)=87.0kJ//mol Delta_(f)G^(oplus)(O_(2))=0kJ//mol
For the reaction, 2A(g)+B(g)to2D(g) DeltaU^(theta)=-10.5 kJ and DeltaS^(theta)=-44.1 JK^(-1) Calculate DeltaG^(theta) for the reaction, and predict whether the reaction can occur spontaneously or not.
For , A + B hArr C. the equilibrium concentration of A and B at a temperäture are 15 mol lit^(-1) When volume is doubled the reaction has equilibrium concentration of A as 10 mol lit^(-1) then
For A+BhArrC , the equilibrium concentrations of A and B at a temperature are 15 "mol" L^(-1) . When volume is doubled the reaction has equilibrium concentration of A is 10 "mol" L^(-1) . Calculate a. K_(c) b concentration of C in original equilibrium.
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column