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postnew [5]
3 years ago
10

The equilibrium: 2 NO2(g) \Longleftrightarrow⇔ N2O4(g) has Kc = 4.7 at 100ºC. What is true about the rates of the forwar

d (ratefor) and reverse (raterev) reactions initially and at equilibrium if an empty container is filled with just NO2?
Initial: forward rate < reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate > reverse rate
Initial: forward rate > reverse rate Equilibrium: forward rate = reverse rate
Initial: forward rate = reverse rate Equilibrium: forward rate = reverse rate
Chemistry
1 answer:
Sergeeva-Olga [200]3 years ago
7 0

Answer:

  • Initial: forward rate > reverse rate
  • Equilibrium: forward rate = reverse rate

Explanation:

2NO₂(g) → N₂O₄(g)   Kc=4.7

The definition of <em>equilibrium</em> is when the forward rate and the reverse rate are <em>equal</em>.

Because in the initial state there's only NO₂, there's no possibility for the reverse reaction (from N₂O₄ to NO₂). Thus the forward rate will be larger than the reverse rate.

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Fiesta28 [93]

Complete Question

Questions Diagram is attached below

Answer:

*  W=1142.86Joule

*  Q=997.7J

*  H=2140.5J

Explanation:

From the question we are told that:

Temperature T=337K

Pressure P=(60-55)Pa*10^5

VolumeV=(1.6-1.4)m^3*10^{-3}

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\frac{P_2}{P_1}=\frac{V_1}{V_2}^n

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Therefore

Work-done

 W=\int{pdv}

 W=\frac{55*10^5*1.6*10^{-3}*60*10^5*1.4*10^{-3}}{1-0.65}

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Generally the equation for internal energy is mathematically given by

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