2SO\(_2\)\(_{(s)}\)  +  O\(_2\)\(_{(s)}\) ⇌  2SO\(_3\) ; ΔG° = - ve

For the above reaction to be feasible

Lord_Morgan

3 months ago

no correct option actually.

maybe due to typographical error.

∆H = -ve , ∆s = -ve

why?

checking the reaction


reactants have 3 moles of gas, product has 2


so entropy has decreased


∆s = -ve


from


∆G = ∆H - T∆S


the only way to get a negative ∆G


is if ∆H is negative and high enough to counter the T∆S

FGCO195928

3 months ago

it's either A or B, but I think the ∆H can't be 0 because that means the temperature of the reaction is 0 and can a reaction take place at 0K abi 0°C

Henry1472

4 months ago

where are the correct options

mustapha2010

4 months ago

no Correct option