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soldier1979 [14.2K]
3 years ago
8

The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia: 2N0(g) +02(g) 2NO2 (g) The

net reaction is: 4NH, (g) [email protected])= 4 Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K1 and K2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator. K=
Chemistry
1 answer:
Tresset [83]3 years ago
4 0

The question is incomplete, complete question is :

The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

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

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols

Answer:

Equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

Explanation:

Step 1 : 4NH_3(g)+5O_2(g)\rightleftharpoons 4NO(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}

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

Expression of an equilibrium constant can be written as:

K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}

The net reaction is:

4NH_3(g)+7O_2(g)\rightleftharpoons 4NO_2(g)+6H_2O(g)

Expression of an equilibrium constant can be written as:

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}

Multiply and divide [NO]^4;

K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}

K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}

K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}

K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2

K=K_1\time (K_2)^2

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 \& K_2:

K=K_1\time (K_2)^2

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Answer:

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Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

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v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

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c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

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e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

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