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Vladimir [108]
3 years ago
13

The equilibrium constant for the reaction H2 (g) + I2 (g) <==> 2 HI (g)

Chemistry
1 answer:
antoniya [11.8K]3 years ago
5 0

The percentage of I2 converted = 78.6 %

Explanation:

Write down the given values

Kp= 54.4

The percentage of I2 will be converted to HI is 0.200 moled each of H2 and I2.

It should come to an equilibrium in 1.00 lit container .

Change in I2 (iodine) and H2 (hydrogen)  = x in each

Change in HI = 2x

total ni (nickel)

number of moles = 0.2 -x + 0.2 -x + 2x

=0.4 moles

Mole fractions :

I2 = 0.2-x / 0.4 H2

=0.2-x  / 0.4 HI

= 2x /0.4

Kp = HI ^2 / H2* I2

= (2x) ^2 / (0.2-x) ^2 = 54.4

by  taking square root:

2x / 0.2-x = 7.375

= x=0.157

percentage of I2 converted = 78.6 %

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An orgo lab student named Bob wanted to synthesize an isoamyl acetate ester because it smells like bananas and he really likes t
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Explanation:

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Isoamyl acetate grams:

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How long would it take Jesse, with an acceleration of -2.50 m/s², to bring his bicycle, with an initial velocity of 13.5 m/s, to
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6 0
2 years ago
A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper
Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

      2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}

Equation for reaction quotient is as follows.

         Q = \frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}

             = \frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}

             = 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of SO_{3} as equilibrium is achieved by using Q, is as follows.

  • This means that there are too much products.
  • Equilibrium will shift to the left towards reactants.
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3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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