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natulia [17]
3 years ago
13

Consider the following reaction at equilibrium. 2CO2 (g) 2CO (g) + O2 (g) H° = -514 kJ Le Châtelier's principle predicts that th

e equilibrium partial pressure of CO (g) can be maximized by carrying out the reaction ________. a. at high temperature and high pressure b. at high temperature and low pressure c. at low temperature and low pressure d. at low temperature and high pressure e. in the presence of solid carbon
Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

C. at low temperature and low pressure.

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 />

  • For the reaction:

<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>

<em></em>

<em><u>Effect of pressure:</u></em>

  • When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
  • The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.

<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>

<em></em>

<u><em>Effect of temperature:</em></u>

  • The reaction is exothermic because the sign of ΔH is (negative).
  • So, we can write the reaction as:

<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>

  • Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>

<em></em>

  • So, the right choice is:

<em>C. at low temperature and low pressure.</em>

<em></em>

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Answer:

4.67M

Explanation:

The concentration of methanol (CH3OH) can be calculated using the following:

Molarity (M) = number of moles(n)/volume(v)

However, mole is not given. It can be obtained by using:

Mole = mass / molar mass

Where; mass = 34.4g

Molar mass (MM) of CH3OH is:

= 12 + 1(3) + 16 + 1

= 12 + 3 + 17

= 32g/mol

mole = 34.4/32

mole = 1.075mol

The volume needs to be converted to L by dividing by 1000

230mL = 230/1000

= 0.230L

Molarity = mol/volume

Molarity = 1.075/0.230

Molarity = 4.6739

Molarity = 4.67M

The concentration of CH3OH in solution is 4.67M

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Definition of chemical change
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The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P
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Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

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To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is \rm PbS. There will be 890 grams of \rm PbS in one kilogram of this rock.

Formula mass of \rm PbS:

M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}.

How many moles of \rm PbS formula units in that 890 grams of \rm PbS?

\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol.

There's one mole of \rm Pb in each mole of \rm PbS. There are thus \rm 3.71980\; mol of \rm Pb in one kilogram of this rock.

What will be the mass of that \rm 3.71980\; mol of \rm Pb?

m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg.

In other words, the \rm PbS in 1 kilogram of this rock contains \rm 0.770743\; kg of lead \rm Pb.

How many kilograms of the rock will contain enough \rm PbS to provide 1.5 kilogram of \rm Pb?

\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg.

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