During a reaction, the enthalpy of formation of an intermediate is 90.3 kJ/mol. During the reaction, 2 moles of the intermediate are formed as a reactant. What is the enthalpy value for this step of the reaction?A.-90.3 kJB.90.3 kJC.180.6 kJD.-180.6 kJSUBMITarrow_backPREVIOUS

Given the following enthalpies of formation in kJ/mol:CH4(g) = - 74.8           O3(g) = 143H2O(g) = - 242             CO2(g) = - 394what is the standard heat of reaction in kJ/mol for the reaction below?3 CH4(g) + 4 O3(g) 3 CO2(g) + 6 H2O(g)Question 6Select one:a.(3 x -394   +  6 x -242)   +   ( 3 x -74.8  +   4 x 143) b.(3 x -394   +  6 x -242)   -   ( 3 x 74.8  +   4 x  -143) c.(3 x -394   +  6 x -242)   -   ( 3 x -74.8  +   4 x 143) d.( 3 x -74.8  +   4 x  143)  -  (3 x -394   +  6 x -242)e.( 3 x -74.8  +   4 x  143)  +  (3 x -394   +  6 x -242)

The enthalpy of a reaction is +270 kJ/mol. What can you determine about the reaction?A.It is exothermic.B.It is nonspontaneous.C.It is endothermic.D.It is spontaneous.

If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem, what would be the final value for the enthalpy of reaction you use for this intermediate reaction?H2 + 0.5 O2 H2O, H = -286 kJA.-286 kJB.286 kJC.572 kJD.-572 kJSUBMITarrow_backPREVIOUS

Given that the reaction,2 SO3(g) -> 2 SO2(g) + O2(g)has an enthalpy, (ΔH) of 198.2 kJ, What would be the enthalpy for the reverse reaction?Group of answer choices-198.2 kJ198.2 kJ

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