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Naily [24]
1 year ago
15

The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C. 2NO (g) + O2 (g) 2NO2 (g) What is the value of Keq at this temperat

ure for the following reaction? NO2 (g) NO (g) + 1/2 O2 (g)
Chemistry
1 answer:
kodGreya [7K]1 year ago
5 0

<u>Answer:</u> The equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

<u>Explanation:</u>

The given chemical equation follows:

2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)

The value of equilibrium constant for the above equation is K_{eq}=5.4\times 10^{13}

Calculating the equilibrium constant for the given equation:

NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)

The value of equilibrium constant for the above equation will be:

K'_{eq}=\frac{1}{\sqrt{K_{eq}}}\\\\K'_{eq}=\frac{1}{\sqrt{5.4\times 10^{13}}}\\\\K'_{eq}=1.36\times 10^{-7}

Hence, the equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

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Answer:

d = 70.5 mm

Explanation:

given,

length of pipe = 305 m

discharge rate = 150 gal/min

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150 gal/min = 150 × 6.30902 ×  10⁻⁵ m³/s

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h = \dfrac{flv^2}{2gd}

h = \dfrac{flv^2}{2gd}

Q = A V

h = \dfrac{fl(\dfrac{Q}{A})^2}{2gd}

h = \dfrac{fl(\dfrac{Q}{\dfrac{\pi}{4}d^2})^2}{2gd}

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f = 0.048 from moody chart using P/D = 0.00015

\dfrac{1}{d^5}= \dfrac{6.1\times \pi^2\times 9.8}{8\times 0.048\times 305\times 0.00946^2}

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Diameter of the pipe is equal to 70.5 mm

7 0
2 years ago
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Answer:

(3) the partial pressure of carbon dioxide above the solution is reduced.

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But as we open the bottle the pressure decreases the excess CO_2 (and air) escapes producing effervescence and partial pressure of CO_2 decreases. ( Because of Henry's Law)  

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Estimate ΔG°rxn for the following reaction at 449.0 K. CH2O(g) + 2 H2(g) → CH4(g) + H2O(g) ΔH°= -94.9 kJ; ΔS°= -224.2 J/K
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This equation relates all three variables, so just plug in all of the given values. Don’t forget to convert deltaS into kJ/K because Gibbs free energy is measured in kJ. The K used in temperature cancels out the K in kJ/K, leaving only kJ as your units. Don’t forget sig figs!

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