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xenn [34]
3 years ago
10

At elevated temperatures, methylisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) → CH3CN (g) At the start of the

experiment, there are 0.200 mol of reactant (CH3NC) and 0 mol of product (CH3CN) in the reaction vessel. After 25 min of reaction, 0.108 mol of reactant (CH3NC) remain. The average rate of decomposition of methyl isonitrile, CH3NC, in this 25 min period is ________ mol/min.
Chemistry
1 answer:
Olegator [25]3 years ago
7 0

Answer:

Average rate of decomposition = 3.68 x 10⁻³ mol/min

Explanation:

CH₃NC (g) → CH₃CN (g)

Average rate of decomposition = [CH₃NC]/Δt

[CH₃NC] = initial [CH₃NC] - final [CH₃NC]

[CH₃NC] = 0.200 mol - 0.108 mol

[CH₃NC] = 0.092 mol

Average rate of decomposition = 0.092 mol / 25 min

Average rate of decomposition = 3.68 x 10⁻³ mol/min

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In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
In need of help very soon please.
NemiM [27]

Answer:

hello, i hope this helps.

Explanation:

1 - group

2 - period

3 - periodic table

4 - family

5 - octet rule

6 - valence electrons

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3 years ago
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sergiy2304 [10]

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5 0
3 years ago
Read 2 more answers
25 points please help
maxonik [38]

Answer : Option (A) Accelerator 2 model has the lowest percentage of energy lost as waste.

Solution : Given,

For Accelerator 1 model,

Input energy = 2078.3 J

Wasted energy = 663.1 J

Output energy = 1415.2 J

For Accelerator 2 model,

Input energy = 7690.0 J

Wasted energy = 2337.5 J

Output energy = 5353.5 J

For Accelerator 3 model,

Input energy = 4061.9 J

Wasted energy = 2259.6 J

Output energy = 1802.3 J

Formula used for lowest percentage of energy lost as waste is:

% energy lost as waste = (Total energy wasted / Total input energy )  ×  100

For Accelerator 1 model,

% energy lost as waste = \frac{663.1}{2078.3}\times100 = 31.90%

For Accelerator 2 model,

% energy lost as waste = \frac{2337.5}{7690.0}\times100 = 30.39%

For Accelerator 3 model,

% energy lost as waste = \frac{2259.6}{4061.9}\times100 = 55.62%

So, we conclude that the Accelerator 2 model has the lowest percentage of energy lost as waste.



6 0
3 years ago
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