If delta H°rxn and delta S°rxn are both positive values, what drives the spontaneous reaction and in what direction at standard conditions?

The spontaneous reaction is
a)enthalpy driven to the left.
b)entropy driven to the right.
c)entropy driven to the left.
d)enthalpy driven to the right.

I can't really picture this in my head because I was thinking if they are both positive then they need to be at high temperature. Then I am stuck. I am confused as what enthalpy and entropy is. I think entropy is delta S then what is enthalpy?

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  1. I'm not exactly sure I understand the question.
    dH is enthalpy. dS is entropy.
    dG = dH - TdS
    So dG is negative for spontaneity.
    I think this is just an algebra exercise but check me out on that.
    If dH is + that's the opposite for getting a negative number for dG so that means the reaction is favored to the left but that isn't spontaneous. I think this is driven to the left so A would be correct, in addition to A, BUT that's not a spontaneous direction.

    B. If dS is + then -TdS is - and means spontaneous. I think this is the answer.

    C. This contradicts A

    D. This contradicts B.

    If not for that spontaneous statement just before the multiple guesses, I would think it could have two answers. That is, +dH does mean it is driven to the left but that doesn't coincide with the preamble to the answers. Sorry if this confuses you.

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  2. Thanks DrBob222!

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  3. Dr Bob is wrong, its goes to the right with the driving force of enthalpy

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