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Chemistry(Please check answers)

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1) For a reaction delta Go is more negative than delta Ho. What does this mean?

a. delta So is zero.
b. delta So is negative
c. delta So is positive
d. delta So is negative if delta Ho is positive.
e. delta So is negative if delta Ho is negative.

I think that delta S is negative if delta H is positive.

2) Sodium reacts violently with water according to the equation:

2 Na (s) + 2 H2O (l) -> 2 NaOH (aq) + H2 (g)

The resulting solution has a higher temperature than the water prior to the addition of the sodium. What are the signs of delta Ho and delta So for this reaction?

a. delta Ho is negative and delta So is positive.
b. delta Ho is positive and delta So is negative.
c. delta Ho and delta So are both negative.
d. delta Ho and delta So are both positive.

I think that the water is endothermic which would make the system exothermic which means that delta H and delta S would be negative.

Would you agree with my answers? Thank you.

  • Chemistry(Please check answers) -

    1. Let's put some numbers in.
    dG = dH - TdS
    -200 = -100 - TdS.
    dG is more negative than dH.
    The only way this can work is for dS to be +. If dS is - then -TdS is + and -100 + any number means dG is not more negative than dH.

    +90 = +100 - TdS
    dG is still more negative than dH.
    If dS is +, then -TdS is - and any - number added 100 makes it smaller so dG < dH.
    If dS is -, then - TdS is + and any + number added to 100 makes dG more positive than dH and not more negative than dH.
    I think the only answer is c.

    2. I agree that water is absorbing heat and that the reaction is exothermic. That means dH must be -. Look at the reaction.
    2Na(s) + 2H2O(l) ==> H2(g) + 2NaOH(aq)
    solid to aqueous. liquid to gas. Doesn't that spell + for S?

  • Chemistry(Please check answers) -

    For question 2 my actually chose that delta H is negative and delta s is positive I just typed the wrong answer in. Is this correct?

  • Chemistry(Please check answers) -

    Yes, a is correct.

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