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2 years ago

Find the mole fraction of Methanol CH3OH and water in a solution prepared by dissolving 4.5 g of alcohol in 40 g of H2O.

1 answer:

6 0

Calculations a technician performs for the CSP should be located on the O a) standard operating procedure (SOP). b) master formulation record (MFR). O) compounding record (CR). d) quality assurance

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3 years ago

Hello, a little help please guys:( Explain how the series of experiments performed by Crookes, Thomson, Rutherford, and Chadwick

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3 years ago

Is a free electron the same as an electron

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3 years ago

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A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of t

creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

2 SO3(g) → 2 SO2(g) + O2(g)

ΔHf° (kJ mol-1) -395.7 -296.8

S° (J K-1 mol-1) 256.8 248.2 205.1

ΔH° = kJ

S° = J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS = ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) = 2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ = 197.8 - T(0.1879)

and the reaction will become spontaneous when the term T(0.1879) becomes greater that 197.8,

0 = 197.8 - 0.1879 T ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

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3 years ago