Answer:
The correct answer is B. It is spontaneous only at low temperatures.
Explanation:
In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
The spontaneity of a reaction is given by the equation:
ΔG = ΔH - TΔS
where:
ΔH: enthalpy variation
T: absolute temperature
ΔS: entropy variation
As the reaction is exothermic, ΔH<0
As the reaction order increases (the reagents are solid and gas and their product is solid), ΔS<0
Therefore, the reaction will be spontaneous when ΔG is negative.
ΔG = ΔH - TΔS
That is, the entropy term must be smaller than the enthalpy term.
Hence, the reaction will be spontaneous only at low temperatures.
Answer:
what's the question? is it A
Answer:
it's a D answer if it's wrong please can you understand my question please
Answer:
we will use the Clausius-Clapeyron equation to estimate the vapour pressures of the boiling ethanol at sea level pressure of 760mmHg:
ln (P2/P1) = ![\frac{ΔvapH}{R}([tex]\frac{1}{T1}](https://tex.z-dn.net/?f=%5Cfrac%7B%CE%94vapH%7D%7BR%7D%28%5Btex%5D%5Cfrac%7B1%7D%7BT1%7D)
-
)
where
P1 and P2 are the vapour pressures at temperatures T1 and T2
Δ
vapH = the enthalpy of vaporization of the ETHANOL
R = the Universal Gas Constant
In this problem,
P
1
=
100 mmHg
; T
1
=
34.7 °C
=
307.07 K
P
2
=
760mmHg
T
2
=T⁻²=?
Δ
vap
H
=
38.6 kJ/mol
R
=
0.008314 kJ⋅K
-1
mol
-1
ln
(
760/10)=(0.00325 - T⁻²) (38.6kJ⋅mol-1
/0.008314
)
0.0004368=(0.00325 - T⁻²)
T⁻²=0.002813
T² = 355.47K
