For the reaction `2C_(6)H_(5)CO_(2)H(s)`+`13O_(2)(g)` gives `12C(g)` + `6H_(2)O(g)`
`U^(0)`=-772.7 `kJmol^(-1)` at 298K Calculate `H^(0)` .
(A) +760.3 `kJmol^(-1)`
(B)-760.3 `kJmol^(-1)`
(C) +670.3 `kJmol^(-1)`
(D) -790.3 `kJmol^(-1)`

For the reaction : C_(2)H_(5)OH(l)+3O_(2)(g)rarr2CO_(2)(g)+3H_(2)O(g) if Delta U^(@)= -1373 kJ mol^(-1) at 298 K . Calculate Delta H^(@)

For the reaction 2\(C_{6}H_{5}CO_{2}\)H(s)+13\(O_{2}(g)\rightarrow 12CO_{2} +H_{2}O\) \(\Delta U^{O} = -772.7kJmol^{-1}\) at 298K . Calculate \(\Delta H^{0}\) . (A) \(+760.3kJmol^{-1}\) (B) \(-760.3kJmol^{-1}\) (C) \(+670.3kJmol^{-1}\) (D) \(-790.3kJmol^{-1}\)

For the reaction CH_(4(g))+2O_(2(g))hArrCO_(2(g))+2H_(2)O_(l) : ( DeltaH=-170.8 kJmol^(-1) ). Which of the following statement is not true?

Calculate the equilibrium constant for the following reaction at 298K and 1 atmospheric pressure: C(graphite) +H_(2)O(l) rarr CO(g) +H_(2)(g) Given Delta_(f)H^(Theta) at 298 K for H_(2)O(l) =- 286.0 kJ mol^(-1) for CO(g) =- 110.5 kJ mol^(-1) DeltaS^(Theta) at 298K for the reaction = 252.6 J K^(-1) mol^(-1)

For the reaction N_(2)(G)+3H_(2)(g)hArr2NH_(3)(g),Delta=-93.6KJmol^(-1) The number of moles o fH_(2) at equilibrium will increase If :