Question

1) Predict the sign, (+) or (-), for ∆H and ∆S for the following reactions.

2 C(s) + O (g)         2CO(g)                                 ∆H: _____                  ∆S: _____

CO₂ (s)      CO₂ (g)                                              ∆H: _____                  ∆S: _____

N₂(g) + 3H₂(g)      2NH₃(g)                                 ∆H: _____                  ∆S: _____

2) Use the standard entropy values below to calculate the entropy change (∆S^o_rxn) for the following balanced equation. N₂ (g) + 2 O₂ (g) --> 2 NO₂ (g)

Species        S^{o , (J/K*mol)}   

N₂(g)            191.5                                

O₂(g)            205.0                                                     

NO₂(g)         240.0

3) A reaction for which ∆H and ∆S are both negative will be:     

a. spontaneous at high temperatures.

b. spontaneous at low temperatures.

c. spontaneous at all temperatures.

d. never be spontaneous

e. more information is needed to determine spontaneity.

Answers I got: 1) - + , + - , + - 2) -121.5 3) I think its A.

Thanks

Answer 1

Q1.

1) Predict the sign, (+) or (-), for ∆H and ∆S for the following reactions.

2 C(s) + O (g)         2CO(g)                                 ∆H: _____                  ∆S: _____

dH must be NEGATIVE, since this is combustion, which releases heat

dS --> entropy must increases since 1 mol of gas forms 2 mol of gases, which is more chaotic

CO2 (s)      CO2 (g)                                              ∆H: _____                  ∆S: _____

This is sublimation, it require energy to go from solid to energy, so this must be positive, endothermic

dS --> increases since gas is much more chaotic than solid

N2(g) + 3H2(g)      2NH3(g)                                 ∆H: _____                  ∆S: _____

This is dH = negative, since bonds are broken and energy is released

dS --> 4 mol of gas form 2 mol of gas, chaos decreases so dS must be negative

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