For hypothetical reversible reaction
`1//2A_(2)(g) +3//2B_(2)(g) rarr AB_(3)(g), DeltaH =- 210 kJ `if standard entropies of `A_(2),B_(2)`, and `AB_(3)` are `60,40` and `50JK^(-1) mol^(-1)` respectively. The above reaction will be equilibrium at

For a reversible reaction : X_((g))+3Y_((g))hArr 2Z_((g)), Delta H=-40 kJ , the standard entropies of X,Y and Z are 60, 40 and 50 JK^(-1)mol^(-1) respectively. The temperature at which the above reaction attains equilibrium is about

For reversible reaction : X_((g))+3Y_((g))hArr 2Z_((g)), DeltaH=-"40 kJ" Standard entropies of X, Y and Z are 60, 40and "50 J K"^(-1)",ol"^(-1) respectively. The temperature at which the above reaction is in equilibrium is

For reversible reaction : X_((g))+3Y_((g))hArr 2Z_((g)), DeltaH=-"40 kJ" Standard entropies of X, Y and Z are 60, 40and "50 J K"^(-1)",ol"^(-1) respectively. The temperature at which the above reaction is in equilibrium is

For the reaction (1)/(2) N_(2)(g) + (3)/(2) H_(2)(g) rarr NH_(3)(g), DeltaH =-30 KJ to be at equilibrium at 477^(@) C. If standard entropy of N_(2)(g) and NH_(3)(g) are 60 and 50 "J" "mole"^(-1)K^(-1) respectively then calculate the standard entropy of H_(2)(g) in "J mole"^(-1) K^(-1).

For the reaction (1)/(2) N_(2)(g) + (3)/(2) H_(2)(g) rarr NH_(3)(g), DeltaH =-30 KJ to be at equilibrium at 477^(@) C. If standard entropy of N_(2)(g) and NH_(3)(g) are 60 and 50 "J" "mole"^(-1)K^(-1) respectively then calculate the standard entropy of H_(2)(g) in "J mole"^(-1) K^(-1).