Vapour pressure of the water at `80^(0)C` is "0.467atm." What is `K_(C)` of the following equilibrium?
`H_(2)O_((l))` `harr` `H_(2)O_((g))`
(A) "0.8959mol/L
(B) `2.9times10^(-4)`mol/L
(c) 0.016mol/L
(D) 13.19mol/L

K_(c) for the reaction N_(2)+3H_(2) hArr 2NH_(3) is 0.5 mol^(-2) L^(2) at 400 K . Find K_(p) . Given R=0.082 L-"atm deg"^(-1) mol^(-1)

At 90^(@)C pure water has [H_(3)O^(+)] as 10^(-6) mol L^(-1) . What is the value of K_(w) at 90^(@)C ?

Calculate the standard internal energy change for the following reactions at 25^(@)C : 2H_(2)O_(2)(l) rarr 2H_(2)O(l) +O_(2)(g) Delta_(f)H^(Theta) at 25^(@)C for H_(2)O_(2)(l) =- 188 kJ mol^(-1) H_(2)O(l) =- 286 kJ mol^(-1)

Calculate the change in enthalpy for the following process at 1 atm H_(2)O(1, 50^(@)C) rarr H_(2)O (g, 150^(@)C) given that Delta H_(v) at 100^(@)C is 40.7 kJ mol^(-1) C_(p)(H_(2)O, l) = 75.0 J mol^(-1)K^(-1) C_(p)(H_(2)O, g) = 33.3 J mol^(-1)K^(-1)

At 90^(@)C , pure water has [H_(3)O^(+)] as 10^(-6) mol L^(-1) . What is the value of K_(w) at 90^(@)C ?

Ice and water are at equilibrium at 0^@ C , DeltaH =4 kJ/mol for the process, H_2 O_((s)) to H_2 O_((l)) which of the following is CORRECT?

What is the value of change in internal energy at 1 atm in the process? H_(2)O(l,323K)rarrH_(2)O(g,423K) Given : C_(v,m)(H_(2)O,l)=75.0JK^(-1) mol^(-1):" "C_(p,m)(H_(2)O,g)=33.314JK^(-1)mol^(-1) DeltaH_(vap)"ar 373 K"=40.7KJ//mol

For the system A(g)+2B(g) hArr C(g) the equilibrium concentration is A=0.06 mol L^(-1), B=0.12 mol L^(-1) C=0.216 mol L^(-1) The K_(eq) for the reaction is

19 gm of ice is converted into water at 0^(@)C and 1 atm. The entropies of H_(2)O(s) and H_(2)O(l) are 38.2 and 60 J//"mol " K respectively. The enthalpy change for this conversion is :

18g of ice is converted into water at 0^(@)C and 1 atm. The entropies of H_(2)O (s) and H_(2)O (l) are 38.2 and 60J/mol K respectively. The enthalpy cange for this conversion is: