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skad [1K]
3 years ago
5

Consider the following reaction at equilibrium: C(s)+H2O(g)⇌CO(g)+H2(g) Predict whether the reaction will shift left, shift righ

t, or remain unchanged upon each of the following disturbances. a) C is added to the reaction mixture. b) H2Ois condensed and removed from the reaction mixture c) CO is added to the reaction mixture d) H2 is removed from the reaction mixture.
Chemistry
1 answer:
vazorg [7]3 years ago
8 0

'Answer:

a) C is added to the reaction mixture.      "shift right"

b) H₂O is condensed and removed from the reaction mixture.    "shift left"

c) CO is added to the reaction mixture.        "shift left"

d) H₂ is removed from the reaction mixture.      "shift right"

Explanation:

  • <em>Le Châtelier's principle</em><em> states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

<em />

<u><em>a) C is added to the reaction mixture:</em></u>

Adding C(s) will increase the concentration of the reactants side, so the reaction will be shifted to the right side to suppress the increase in the concentration of C by addition.

<em>So, the reaction shift right.</em>

<em></em>

<u><em>b) H₂O is condensed and removed from the reaction mixture:</em></u>

Removing H₂O will decrease the concentration of the reactants side, so the reaction will be shifted to the left side to suppress the decrease in the concentration of H₂O by removal.

<em>So, the reaction shift left.</em>

<em></em>

<u><em>c) CO is added to the reaction mixture:</em></u>

Adding CO will increase the concentration of the products side, so the reaction will be shifted to the left side to suppress the increase in the concentration of CO by addition.

<em>So, the reaction shift left.</em>

<em></em>

<u><em>d) H₂ is removed from the reaction mixture:</em></u>

Removing H₂ will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the increase in the concentration of H₂ by removal.

So, the reaction shift right.

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Explanation:

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When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample
andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

Q=m\times c\times \Delta T

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

\Delta T = change in temperature = 1.643^oC

Now put all the given value in the above formula, we get:

26380J/g=0.235g\times c\times 1.643^oC

c=68323.38J/^oC

Thus, the specific heat of calorimeter is 68323.38J/^oC

Now we have to calculate the heat of combustion of caffeine.

Formula used :

Q=c\times \Delta T

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = 68323.38J/^oC

\Delta T = change in temperature = 1.584^oC

Now put all the given value in the above formula, we get:

Q=68323.38J/^oC\times 1.584^oC

Q=108224.23J=108.2kJ

Now we have to calculate the moles of caffeine.

\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

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3 years ago
A gas occupies 1200 litres at 2 atm pressure. To what pressure must it be compressed to occupy 60 litres at the same temperature
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P2 = 40 atm

Explanation:

Given:

P1 = 2 atm

V1 = 1200 L

V2 = 60 L

P2 = ?

Using Boyle's law and solving for P2,

P1V1 = P2V2

P2 = (V1/V2)P1

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Calculate the percent yield if 52.9 g of ethanol reacts to produce 26.7 g of ether.
blagie [28]

Answer:

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Explanation:

<u>Step 1:</u> the balanced equation

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<u>Step 2:</u> Data given

mass of ethanol = 52.9 grams

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mass of ether = 26.7 grams

Molar mass of ether = 74.12 g/mol

<u>Step 3:</u> Calculate moles of ethanol

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For 1.15 moles of ethanol consumed, we have 1.15/2 = 0.575 moles of ether produced

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% yield = (26.7 / 42.62) * 100 = 62.7%

The % yield is 62.7 %

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