Chemical energy
the energy stored in the bonds of atoms and molecules
ΔG deg will be negative above 7.27e+3 K.
<u>Explanation:</u>
- The ΔG deg with the temperature can be found using the formula and the formula is given below
- ΔG deg = ΔH deg - T ΔS deg
- Given data, ΔH deg = 181kJ and ΔSdeg=24.9J/K
- -T ΔS deg will be always negative and ΔG deg = ΔH deg will be positive and ΔG deg will be negative at relatively high temperatures and positive at relatively low temperatures
- solving the equation and substitute ΔGdeg=0
- ΔGdeg = ΔHdeg - T ΔSdeg
- T= ΔHdeg/ΔSdeg
- T=181 kJ / 2.49e-2 kJK-1
- By simplification we get
- T=7.27 × 10^3 K.
- Therefore, Go will be negative above 7.27 × 10^3 K
- Since ΔG deg = -RT lnK, when ΔGdeg < 0, K > 1 so the reaction will have K > 1 above 7.27 × 10^3 K.
- ΔG deg will be negative above 7.27e+3 K.
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Answer:
Cr₂O₇⁻²(aq) and ClO₃⁻(aq)
Explanation:
At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.
The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:
RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).
Ni⁺²(aq) E° = -0.257 V
I₂(s) E° = +0.535 V
Cr₂O₇⁻²(aq) E° = +1.33 V
ClO₃⁻(aq) E° = +0.890 V
Pb²⁺(aq) E° = -0.125 V
So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).
a. pressure increases
Gay Lussac's Law
When the volume constant, the gas pressure is proportional to its absolute temperature