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rosijanka [135]
3 years ago
13

Iron(III) oxide can be reduced by carbon monoxide:Fe2O3 (s) + 3 CO (g) ⇌ 2 Fe (s) + 3 CO2 (g) Use the following thermodynamic da

ta at 298 K to determine the equilibrium constant at this temperature.SubstanceFe2O3 (s)CO (g)Fe (s)CO2 (g) ∆Hfº(kJ /mol)–824.2–110.50 –393.5∆Gfº(kJ/mol)–742.2 –137.20 –394.4Sº (J/mol•K)87.4 197.727.78213.7
Chemistry
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

k = 1,423x10⁵

Explanation:

For the reaction:

Fe₂O₃(s) + 3CO (g) ⇌ 2Fe(s) + 3CO₂(g)

You can obtain equilibrium constant, k, using:

ΔG° = -RT lnK <em>(1)</em>

<em></em>

ΔG° of the reaction is:

ΔGf° products - ΔGf° reactants, that means:

ΔG° = 2∆Gfº Fe(s) + 3∆Gfº CO₂(g) - (∆Gfº Fe₂O₃(s) + 3∆Gfº CO (g)

ΔG° = 2×0 + 3×-394,4kJ/mol - (-742,2kJ/mol + 3×-137,2kJ/mol)

<em>ΔG° = -29,4 kJ/mol</em>

<em />

Replacing in (1) knowing R = 8,314472x10⁻³ kJ/molK; T = 298K

-29,4 kJ/mol = -8,314472x10⁻³ kJ/molK 298K lnK

11,866 = lnK

<em>1,423x10⁵ = k </em>

<em></em>

I hope it helps!

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reddi tstudent has a thin copper beaker containing 100 g of a pure metal in the solid state. The metal is at 215°C, its exact me
prohojiy [21]

Explanation:

A point of temperature at which both solid and liquid state of a substance remains in equilibrium without any change in temperature then this temperature is known as melting point.

For example, melting point of water is 0 ^{o}C. So, at this temperature solid state of water and liquid state are present in equilibrium with each other.

Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is 215^{o}C then some of the solid will change into liquid state but the temperature will remains the same.  

4 0
3 years ago
How many of the following are oxidation-reduction reactions? I. reaction of a metal with a nonmetal II. synthesis III. combustio
nadezda [96]

Answer : The oxidation-reduction reactions are:

I. reaction of a metal with a nonmetal

II. synthesis

III. combustion

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

I. Reaction of a metal with a nonmetal :

When sodium react with chlorine gas then it react to give sodium chloride.

2Na+Cl_2\rightarrow 2NaCl

In this reaction, the oxidation state of sodium changes from (0) to (+1) and shows oxidation and the oxidation state of chlorine changes from (0) to (-1) and shows reduction. So, It is an oxidation-reduction reaction.

II. Synthesis reaction :

A chemical reaction where multiple substances or reactants combine to form a single product.

When hydrogen react with oxygen then it react to give water.

2H_2+O_2\rightarrow 2H_2O

In this reaction, the oxidation state of hydrogen changes from (0) to (+1) and shows oxidation and the oxidation state of oxygen changes from (0) to (-2) and shows reduction. So, It is an oxidation-reduction reaction.

III. Combustion reaction :

A chemical reaction in which a hydrocarbon reaction with the oxygen to give product as carbon dioxide and water.

When methane react with oxygen gas then it react to give carbon dioxide and water.

CH_4+2O_2\rightarrow CO_2+2H_2O

In this reaction, the carbon of methane gain oxygen and shows oxidation and the oxygen gas gain hydrogen and shows reduction. So, It is an oxidation-reduction reaction.

IV. Precipitation reaction :

It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

It is a double displacement reaction. So, it is not an oxidation-reduction reaction.

V. Decomposition reaction :

A chemical reaction in which the the larger molecule decomposes to give two or more smaller molecules.

The oxidation state remains same on reactant and product side. So, it is not an oxidation-reduction reaction.

3 0
3 years ago
A textbook measures 250 mm long, 225 mm wide and 50 mm thick. What is the volume of this book in mm3? What is the volume of this
Dvinal [7]

Answer:

2.81 × 10⁶ mm³

2.81 × 10⁻³ m³

Explanation:

Step 1: Given data

Length (l): 250 mm

Width (w): 225 mm

Thickness (t): 50 mm

Step 2: Calculate the volume of the textbook

The book is a cuboid so we can find its volume (V) using the following expression.

V = l × w × t = 250 mm × 225 mm × 50 mm = 2.81 × 10⁶ mm³

Step 3: Convert the volume to cubic meters

We will use the relationship 1 m³ = 10⁹ mm³.

2.81 × 10⁶ mm³ × 1 m³ / 10⁹ mm³ = 2.81 × 10⁻³ m³

6 0
3 years ago
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol
Kryger [21]

Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant K_c=  \dfrac{[CO][H_2]}{[H_2O]}

The equilibrium constant  K_c=  \dfrac{(0.17 )(0.17)}{0.74}

The equilibrium constant K_c=  0.03905

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \  \ \ \ \ \to0.17

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

K_c=  \dfrac{[CO][H_2]}{[H_2O]}

0.03905 =  \dfrac{[0.17+x][x]}{[0.91 -x]}

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265  or   x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

3 0
3 years ago
KOH(Ac) + HNO3 → KNO3(Ac) + H2O<br> cual es su reacio
Bumek [7]

Answer:

corbonization for dictionnal

Explanation:

correct

6 0
2 years ago
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