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nignag [31]
1 year ago
5

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0 °C for the following reacti

on. 2 NOCI(g) 2 NO(g) + Cl2(g) Round your answer to 2 significant digits. K= 0 х $ ? For a certain chemical reaction, the standard Gibbs free energy of reaction at 25.0°C is 51.4 kJ. Calculate the equilibrium constant K for this reaction Round your answer to 2 significant digits.
Physics
1 answer:
Yakvenalex [24]1 year ago
6 0

The equilibrium constant of the reaction at 25⁰c will be 426827.5.

<u />

  • <u>Theory-</u>

<u>Equilibrium constant</u> :The equilibrium constant comes from the chemical equilibrium law. For the chemical equilibrium state, at a fixed constant temperature, the ratio of the product of the reaction's multiplication to the concentration of its reactants' multiplication, and each is raised to the power to the corresponding coefficients of the elements in the reaction.

The chemical equilibrium is given by for a general chemical reaction.

a. A+ b. B ⇌ c. C+ d. D,.

Kc =[C]c [D]d/[A]a [B]b.

<u>Gibb's free energy</u> :The second law of thermodynamics can be arranged in such a way that it gives a new expression when a chemical reaction happens at a constant temperature and constant pressure.

G=H-TS

  • <u>Calculations</u>:-

T=25⁰c

G=51.4 x 10³J

\\\\k=GR+\frac{nRT}{Z} \\

k= equilibrium constant ,G=Gibbs free energy ,n= no. of moles ,R=Gas constant ,T=temperature ,Z=compressibility

Ideal.Situation=\left \{ {{Z=1} \atop {n=1}} \right.

\\\\\\k=GR+RT

k=51.4 x 10³ x 8.3 + 8.3 x 25

k=426827.5

To learn equilibrium constant-

<u>brainly.com/question/19669218</u>

#SPJ4

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To find work, you use the equation: W = Force X Distance X Cos (0 degrees)
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So you would do 75 N x 10m x Cos (0 degrees)= 750 J
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2 years ago
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A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in
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Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

v_{av}=\frac{d}{t}

where,

v_{av} = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

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2 years ago
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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.23 2.23 times a second. A tack is stuck in the t
elena-14-01-66 [18.8K]

Explanation:

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       Angular velocity (\omega) = 2.23 rps

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First, we will convert revolutions per second into radian per second as follows.

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Thus, we can conclude that the tack's tangential speed is 5.31 m/s.

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