Help please 20points
2 answers:
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Answer:
a) The slope of the line of best fit plot = -12629.507
b) ΔH∘ = 105 kJ
c) Intercept of the line of best fit plot = 39.099
d) ΔS∘ = 325.1 J/K
e) Option A is correct.
Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.
f) Option D is correct. All of the options are correct.
Explanation:
The complete question is presented in the first attached image to this question. This complete question has the data readings required to plot the graph.
The second attached image has the plotted graph and the regression analysis to obtain the line of best fit.
The equation of the line of best fit obtained is
y = -12629.507x + 39.099
Comparing the given expression for the question with the equation of a straight line
ln (K) = (−ΔH∘/RT) + (ΔS∘/R)
y = mx + c
y = In K
Slope = m = (−ΔH∘/R)
x = (1/T)
Intercept = c = (ΔS∘/R)
So, to answer the question now
a) The slope of the line of best fit plot = -12629.507
b) Slope = (−ΔH∘/R)
(−ΔH∘/R) = -12629.507
But R = molar gas constant = 8.314 J/mol.K
ΔH∘ = 12629.507 × 8.314 = 105,001.721198 J = 105,002 J = 105 kJ
c) Intercept of the line of best fit plot = 39.099
d) Intercept = (ΔS∘/R)
(ΔS∘/R) = 39.099
ΔS∘ = 39.099 × 8.314 = 325.069086 J/K = 325.1 J/K
e) Do you expect the solubility of Borax to increase or decrease as temperature increases?
Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.
f) Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why.
A. To make sure no more sodium borate would dissolve in solution.
B. To ensure the dissolution process was at equilibrium.
C. To make sure the solution was saturated with sodium and borate ions.
D. All of the above
Hope this Helps!!!
Answer:
The change in entropy is -1083.112 joules per kilogram-Kelvin.
Explanation:
If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (), in joules per gram-Kelvin, by the following model:
(1)
Where:
- Mass, in kilograms.
- Specific heat of water, in joules per kilogram-Kelvin.
,
- Initial and final temperatures of water, in Kelvin.
If we know that ,
,
and
, then the change in entropy for the entire process is:
The change in entropy is -1083.112 joules per kilogram-Kelvin.