The formula for the change in Gibbs energy of a solid is:
ΔG = Vm ΔP
where, ΔG is change in Gibbs, Vm is molar volume, ΔP is
change in pressure
ΔP = P(final) – P(initial)
P(final) = 1 atm = 101325 Pa
P(initial) = ρ_water *g *h = (1030 kg/m^3) * 9.8 m/s^2 *
2000 m = 20188000 kg m/s^2 = 20188000 Pa
Vm = (950 kg/m^3) * (1000 mol / 891.48 kg) = 1065.64
mol/m^3
So,
ΔG = (1065.64 mol/m^3) * (101325 Pa - 20188000 Pa)
<span>ΔG = -21405164347 J = -21.4 GJ</span>
The answer is Latitude (B)
Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32
Colligative properties are those substances that depend
on the number of substances in the solution, not in the identity of that
substance. The property changes the way that it does when the amount of solute
is increased because it enables the solute to be scattered more. For example,
the freezing point of salt water is lower than that of the pure water due to
the salt ions present in water.