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Luba_88 [7]
3 years ago
15

A key step in the extraction of iron from its ore is feo(s) + co(g) fe(s) + co2(g) kp = 0.403 at 1000°c. this step occurs in the

700°c to 1200°c zone within a blast furnace. what are the equilibrium partial pressures of co(g) and co2(g) when 1.58 atm co(g) and excess feo(s) react in a sealed container at 1000°c?
Chemistry
1 answer:
Kaylis [27]3 years ago
7 0
The complete reaction of the problem, for better illustration, is

FeO(s) + CO(g) <--> Fe(s) + CO2(g)

The double-tailed arrow signifies that the reaction is in a dynamic chemical equilibrium. When the system is in equilibrium, the forward and the backward reaction rates have an equal ratio of Kp = 0.403 at 1000°C. The formula for Kp is

Kp = [partial pressure of products]/[partial pressure of reactants]

So, first, let's find the partial pressure of the compounds in the reaction.

                      FeO(s) + CO(g) <--> Fe(s) + CO2(g)
Initial                 x            1.58            0           0
Change           -1.58        -1.58       +1.58     +1.58
------------------------------------------------------------------
Equilbrium      x-1.58          0             1.58       1.58

Kp = [(1.58)(1.58)]/[(x-1.58)] = 0.403
x = 7.77 atm (this is the amount of excess FeO)

Therefore, the partial pressure of CO2 at equilibrium is 1.58 atm. There is no more CO because it has been consumed due to excess FeO.
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KengaRu [80]
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6 0
3 years ago
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2 NO ( g
enyata [817]

Answer:

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

Explanation:

2 NO (g)  +  O₂  (g)   ⇄  2NO₂ (g)

Let's apply the thermodynamic formula to calculate the ΔG

ΔG = ΔG° + R .T . lnQ

We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q

How can we know Q? By the partial pressures (Qp)

P NO = 0.450atm

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PNO₂ = 0.650 atm

Qp = [NO₂]² / [NO]² . [O₂]

Qp = 0.650² / 0.450² . 0.1 = 20.86

ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86

ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)

5 0
3 years ago
Calculate the concentration (in mol/l) of 33% by weight (33 g naoh per 100 g of solution) naoh solution. (the density of the 33%
Shtirlitz [24]
Molarity is measured in moles per Liter. If there are 1.35 g/mL, find out how many grams there are in a liter of solution.

If there are 1000 mL in one liter, we can multiply by 1000 to get g/L

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By weight, the NaOH is 33% or .33

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445.5 g  x 1 mol NaOH/39.997 g = 11.13833538 moles per Liter

Rounded to significant figures, the answer is 11 mol/L NaOH


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