51. What is the ΔH° at 298 K of the given reaction?*- 3.8 kJ/mol3.8 J/mol0Cannot defined

Calculate ΔG° for the reaction below under standard conditions at 25°C.  2S(s) + 3O2(g) + 2H2O(l) → 2H2SO4(l) ΔH° = -1056 kJ/molΔS° = -505 J/molHint: Pay attention to units of ΔH° and ΔS°

Consider the following reaction at 298 K.32O2( g)⇌O3( g)⋅KP=2.47×10−29.ΔrG⊕ for the reaction is _______ kJ. (Given R =8.314JK−1 mol−1)

The value of log K for the reaction A⇌B at 298 K is _______. (Nearest integer)Given: ΔH°=−54.07 kJ mol−1ΔS°=10JK−1 mol−1(Taken 2.303×8.314×298=5705 )

The value of ΔH° for the reaction below is - 790 kJ.2S(s) + 3O2(g) → 2SO3(g)The enthalpy change accompanying the reaction of 0.95 g of S is ________kJ.Question 4Select one:a.23b.-23c.-12d.12e.-790

What is ΔrG° (in kJ mol-1) for a redox reaction that has E°cell = -2.69 V at 25°C ? Assume 4 electrons are transferred in the balanced redox reaction.If needed, F = 96485 C/mol.Answer:Question 4