Steam undergoes decomposition at high temperature as per the reaction : `H_(2)O(g)to H_(2)(g)+(1)/(2)O_(2)(g)` `Delta H^(@)=200kJmol^(-1)Delta S^(@)=40JK^(-1)mol^(-1)` The temperature at which equilibrium constant is unity is :

For the reaction Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g) the value of Delta H=30.56 KJ mol^(_1) and Delta S = 66 JK^(-1)mol^(-1) . The temperature at which the free energy change for the reaction will be zero is :-

For a hypothetical reaction A(g) + 3B(g) to 2C(g). Delta H = -100 kJ and Delta S = -200 Jk^(-1) . Then the temperature at which the reaction will be in equilibrium is

The temperature at which the given reaction is at equilibrium Ag_(2)O_(s) rightarrow 2Ag(s) + 1/2 O_(2)(g) Delta H = 40.5 kJ mol^(-1) and DeltaS= 0.086 k J mol^(-1) K^(-1)

The temperature at which the reaction : Ag_(2)O(s)rarr 2Ag(s)+1//2O_(2)(g) is at equilibrium is …….., Given Delta H=30.5 KJ mol^(-1) and Delta S = 0.5 KJ mol^(-1) :

Delta H and Delta S for the reaction, Ag_(2)O(s)to 2A(s)+(1)/(2)O_(2)(g) , are 30.56 kJ mol^(-1) and 66.0 J mol^(-1) respectively. Calculate the temperature at which this reaction will be at equilibrium. Predict whether the forward reaction will be favoured above or below this temperature.

For the transformation, H_(2)O(l,1 atm) "to H_(2)O (g,1atm), DeltaH_(vap)=40.668kJmol^(-) The change in entropy (JK^(-)mol^(-)) is:

Write the equilibrium constant of the reaction C(s)+H_(2)O(g)hArrCO(g)+H_(2)(g)