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kakasveta [241]
3 years ago
6

The most important commercial process for generating hydrogen gas is the water-gas shift reaction:. CH4(g) + H2O(g) → CO(g) + 3H

2(g). Use tabulated thermodynamic data to find ΔG° for this reaction at the standard temperature of 25°C. i found this to be. =1.423×102 kJ. Now calculate ΔG°1200 for this process when it occurs at 1200 K
Chemistry
2 answers:
Sophie [7]3 years ago
7 0
The Gibb's free energy at a certain temperature is calculated by using the equation,
ΔG = - RT(ln K)

where K is the constant. We calculate this value by using the first conditions,
1.423x10^2 = -(8.314)(25 + 273) x (ln K)

Solving for K,
 K = 0.94418

Use the equation for the second set of conditions,

ΔG = - (8.314)(1200) x (ln 0.94418)

<span>ΔG° at 1200 K is approximately 573.05 kJ</span>
PtichkaEL [24]3 years ago
4 0
The Gibb's free energy at a certain temperature is calculated through the equation,
                                 ΔG = - RT(ln K)
where K is constant. Substituting the values from the first set of the given,
                               1.423x10^2 = -(8.314)(25 + 273) x (ln K)
Solving for the value of K gives us an answer of, K = 0.94418
Use the same equation to get the value of ΔG°1200
                                     ΔG = - (8.314)(1200) x (ln 0.94418)
The value of ΔG°1200 is approximately 573.05 kJ

                 
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Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO. The following reaction becomes possible: NO3gNOg 2NO2g The
never [62]

There is an error in the first sentence of the  question; the right format is:

Suppose a 500.mL flask is filled with 1.9mol of NO3 and 1.6mol of NO2.

It should be NO2 and not NO.

Answer:

The equilibrium molarity of NO = 0.21695 m

Explanation:

Given that :

the volume = 500 mL = 0.500 m

number of moles of NO_3 = 1.9 \ mol

number of moles of NO_2 = 1.6 \ mol

Then we can calculate for their respectively concentrations as :

[NO_3] = \frac{number \ of \ moles}{volume}

[NO_3] = \frac{1.9}{0.500}

[NO_3] = 3.8 \ M

[NO_2] = \frac{number \ of \ moles}{volume}

[NO_2] = \frac{}{} \frac{1.6}{0.500}

[NO_2] = 3.2 \ M

The chemical reaction can be written as:

NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}

The ICE table is as follows;

                    NO_3_{(g)} + NO_{(g)} \to 2NO_2_{(g)}

Initial              3.8         -               3.2

Change          +x          x               -2x

Equilibrium     3.8+x    +x              3.2 - 2x

K_c=\frac{[NO_2]^2}{[NO_3][NO]}

where \ K_c = 8.33

8.33 = \frac{(3.2-2x)^2}{(3.8+2x)x} \\ \\ 8.33 = \frac{(3.2-2x)^2}{(3.8x+2x^2)}

8.33(3.8x + 2x^2) = (3.2-2x)^2 \\ \\ 31.654x + 16.66x^2 = (3.2-2x)(3.2-2x) \\ \\ 31.654x + 16.66x^2 = 10.24 - 12.8x +4x^2 \\ \\ 10.24 - 44.454x -12.66 x^2 = 0 \\ \\ 12.66x^2 +44.454x -10.24 = 0

Using quadratic formula;

\frac{-b\pm \sqrt{(b)^2-4ac} }{2a}

= \frac{-(44.454) + \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)} \ \ OR \ \ \frac{-(44.454) - \sqrt{(44.454)^2-4(12.66)(-10.24)} }{2(12.66)}

= 0.21695 OR -3.7283

Going by the positive value;

x = 0.21695

[NO_3] = 3.8 +x  = 3.8 + 0.21695

= 4.01695 m

[NO] = x  = 0.21695 m

[NO_2] = 3.2 +x  = 3.2 + 0.21695

= 3.41695 m

3 0
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Answer:

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The atoms of carbon and phosphorus are most likely to gain electrons from the given choices .

The reason for this is because, both carbon and phosphorus are non-metals. Most non-metals usually accept electrons.

Metals are usually electron donors .

  • Metals are known for their electropositivity which is their ability to lose electrons.
  • Non-metals are electronegative and will tend to have a strong affinity for electrons.
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tresset_1 [31]

Answer:

no

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