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natka813 [3]
3 years ago
9

For each part below, use the data below to calculate ΔG at 215 K for each of the following reactions. Then use the data in the c

hart to determine whether each reaction is spontaneous at 215 K.

Chemistry
1 answer:
ICE Princess25 [194]3 years ago
5 0

Answer:

The first reaction is spontaneous

The second reaction is also spontaneous

Explanation:

The question is incomplete or the data part are missing, nonetheless, here is the complete question with the data table ; For each part below , use the data below to calculate ΔG at 215 K for each of the following reactions . Then use the data in the chart to determine whether each reaction is spontaneous at 215 K.

The Gibb's free energy also referred to as the Gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard Gibb's free energy on the other hand is a state function represented as ΔG, as it depends on the initial and final states of the system. The spontaneity of a reaction is explained by the standard Gibb's free energy.

If ΔG = -ve ( the reaction is spontaneous)

if ΔG = +ve ( the reaction is non-spontaneous)

if ΔG= 0 ( the reaction is at equilibrium)

Use this hints for any reaction involving the Gibb's free, Enthalpy and entropy.

The step by step calculation with the data table is attached.

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2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vapori
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Answer:

The value of W is 5.602 kJ, Q is -70.6 kJ, change in U is -65 kJ, and change in H is -70.3 kJ.

Explanation:

Based on the given information, the mass of CH3OH given is 64 grams, which is condensed isothermally and reversibly to liquid at 64 degrees C. The given standard enthalpy of vaporization of methanol at 64 degrees C is 35.3 kJ per mole.

The moles of CH3OH can be determined by using the formula,  

Moles = Mass / Molar mass

= 64.0 grams / 32.0 grams per mole

= 2 mol

The amount of energy given by the process of condensation is,  

ΔH = 2 mol × 35.3 kJ/mol = 70.6 kJ

In condensation heat is given off, thus, it is an exothermic process, hence, q will be -70.6 kJ

The work or W can be calculated by using the formula,  

W = -P ΔV

Let us first find the volume of 2.0 mole gas at 64 °C, or 64 + 273 = 337 K,  

PV = nRT

V = nRT/P

= 2 mol × 0.08206 L atm per mol K × 337 K/1 atm

= 55.3 L

As the liquid condenses in the process, the change in volume would be negligible. So, the volume change will be -55.3 L

W = - 1 atm × - 55.3 L

W = 55.3 L.atm

W = 55.3 L.atm × 101.3 J/1 L atm = 5602 J

W = 5602 × 1 kJ / 1000 J = 5.602 kJ

W = 5.602 kJ

Now U can be calculated using the formula,  

U = q + W

= -70.6 kJ + 5.602 kJ

= -65. kJ

Thus, q = -70.6 kJ, W = 5.602 kJ, U = -65 kJ, and ΔH = -70.3 kJ.  

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3 years ago
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