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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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A phosphate buffer solution (25.00 mL sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The compone
AysviL [449]

Answer:

0,07448M of phosphate buffer

Explanation:

sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:

Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>

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The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:

0,01862L of HCl×\frac{0,1000mol}{L}= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.

The concentration is:

\frac{1,862x10^{-3}moles}{0,02500L} = <em>0,07448M of phosphate buffer</em>

<em></em>

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