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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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Answer:

W_x = 0.9156\ c

Explanation:

given,

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W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

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5 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
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Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

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We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

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We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

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L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

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