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Bond [772]
2 years ago
9

The irreversible elementary gas-phase reaction is carried out isothermally at 305 K in a packed-bed reactor with 100 kg of catal

yst. The entering pressure was 20 atm and the exit pressure is 2 atm. The feed is equal molar in A and B and the flow is in the turbulent flow regime, with FA0 10 mol/min and CA0 0.4 mol/dm3. Currently 80% conversion is achieved. What would be the conversion
Chemistry
1 answer:
tamaranim1 [39]2 years ago
5 0

Answer:

0.856.

Explanation:

Lets represent the irreversible elementary gas phase equation of reaction as

A + B -----------------------------------> C + D

We have that the percentage of conversion is 80%.

The pressure, p from the ratio of exit pressure and entering pressure is p = 2/20 = 1/10 = 0.1.

Therefore, n = 1 - p^2/ weight of the catalyst = 1 - 0.1^2/ 100 = 9.9 × 10^-3 kg cat^-.

Now, let's make use of the equation below;

J/ 1 - J = kb^2/ u [ w - nw^2/2] ----------(1).

0.8 / 1- 0.8 = k ( 0.4)^2/ 10 [ 100 - (9.9× 10^-3 × 100^2/ 2] .

k = 4.95 dm^6/ kg.cat .mol.min

The turbulent flow= 1/2 × 9.9 × 10^-3 = 4.95 × 10^-3 kg cat^-.

Thus, making use of the equation (1) again, we have that;

{4.95 × 10^-3 × 0.4}/ 10 × [ 100 - (4.95 × 10^-3 × 100^2)] / 2 = 5.964.

Therefore, a/1 - a = 5.964.

5.964( 1 - a) = a.

5.964 - 5.964a = a.

5.964 = a + 5.964a.

5.964 = 6.964a.

a = 5.964/ 6.964 = 0.856.

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Answer:

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Explanation:

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V₁ = 2.0 L                          V₂ = ? L

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(3.2 atm)(2.0 L) = (1.0 atm)V₂                        <----- Insert values

6.4 = (1.0 atm)V₂                                           <----- Simplify left side

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What are the n, l, and possible ml values for the 2p and 5f sublevels?
elixir [45]

Answer:

1. 2p sublevels, n = 2, orbital <em>p</em>, l = 1, ml = 0, ±1

2. 5f sublevels n = 5, orbital <em>f</em>, l = 3, ml = 0, ±1, ±2, ±3

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

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n = 2, orbital <em>p</em>, so l = 1, ml = 0, ±1

2. 5f sublevels

n = 5, orbital <em>f</em>, so l = 3, ml = 0, ±1, ±2, ±3

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