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2 years ago

Enthalpy of CH4(g) + 2NO2(g) -> N2(g) + CO2(g) + 2H2O(l)

Chemistry

1 answer:

stira [4]2 years ago

5 0

Answer:

-177.9 kJ.

Explanation:

Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.

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At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

3 years ago

Calculate the molecular weight of freon-12, if 122.70 liters of its vapor at 200.00

ICE Princess25 [194]

PV = nRT

Where:

P = pressure in atm = 700/760 = 0.9211atm

V = volume = 8.29L

R = gas constant, 0.08206 atm-L/mol-K

T = temperature in Kelvin = 200 + 273 = 473

n = numbers of moles = Mass/molar mass

mass of the compound = 30.5

we can rewrite the equation above as

PV = (Mass)/(Molar mass) * RT

7 0

3 years ago

500ml of a buffer solution contains 0.050 mol nahso3 and 0.031

nydimaria [60]

Answer:

The answers are explained below

Explanation:

Given: concentration of salt/base = 0.031

concentration of acid = 0.050

we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59

we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O

Moles i............0.05...................0.01.................0.031.....................0

Moles r...........-0.01.................-0.01................0.01........................0.01

moles f...........0.04....................0....................0.041.....................0.01

we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041

Hence, we have

PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71

pOH = -log (0.01/0.510) = 1.71

pH = 14 - 1.71 = 12.29

Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.

4 0

3 years ago

1. For each of the following, convert the word equation into a formula equation, BUT do not balance! (4 pts each = 12 pts)

daser333 [38]

Answer:

1a) BaClO₃(s) → BaCl₂(g) + O₂(g)

1b) Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

1c) Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

2a) Calcium hydroxide and hydrogen gas

2b) Tin(II) silicate and Lead(IV) permanganate

2c) Magnesium oxide and water

2d) No product

2e) Mercury and iodine

2f) Calcium chloride and iodine

2g) Strontium phosphite and cesium nitride

2h) Carbon dioxide, water, and sulfur dioxide

2i) Iron oxide(III) and carbon dioxide

2j) Magnesium acetate and hydrogen gas

2k) Calcium iodide

Explanation:

1. For each of the following, convert the word equation into a formula equation, BUT do not balance!

a) Barium chlorate → Barium chloride + Oxygen

<u>1. Chemical formulas</u>

Barium chlorate:

It is a salt: an ionic compound.
Barium has oxidation state +2
Chlorate is the ion ClO₃⁻
Swap the oxidation numbers to write the subscripts: 2 goes to ClO₃ and 1 goes to Ba
Chemical formula Ba(ClO₃)₂
It is solid: Ba(ClO₃)₂(s)

Barium chloride:

It is a salt: an ionic compount
Barium has oxidation state +2
Chlorine is in oxidation state -1
Swap the numbers to write the subscripts: 2 goes to Cl and 1 goes to Ba
BaCl₂
It is solid BaCl₂(s)

Oxygen:

It is a diatomic gas molecule
O₂(g)

<u />

<u>2. Write the unbalanced molecular equation:</u>

BaClO₃(s) → BaCl₂(s) + O₂(g)

b) Chlorine + Potassium nitride → Nitrogen + Potassium chloride

<u>1. Chemical formulas</u>

Chlorine:

It is a diatomic gas molecule
Cl₂(g)

Potassum nitride

It is a salt

Potassium has oxidation state +1
Nitrogen is with oxidation state +3
Swap the oxidation states
K₃N
It is solid: K₃N(s)

Nitrogen:

It is a diatomic gas
N₂(g)

Potassium chloride

It is a salt (ionic compound)
Potassium has oxidation state +1
Chlorine is in oxidation state -1
Swap the oxidation numbers
KCl
It is solid: KCl(s)

<u>2. Write the unbalanced molecular equation</u>

<u />

Cl₂(g) + K₃N(s) → N₂(g) + KCl(s)

c) Sodium nitride + Aluminum bromate → Aluminum nitride + Sodium bromate

<u>1. Chemical formulas</u>

Sodium nitride

It is a salt (ionic compound)
Sodium has oxidation state +1
Nitrogen is with oxidation state -3
Swap the oxidation numbers
Na₃N
It is in aqueous solution
Na₃N (aq)

Aluminum bromate

Salt
Aluminum has oxidation state +3
Bromate is the ion BrO₃⁻
Swap the oxidation states
Al(BrO₃)₃ (aq)

Aluminum nitride

Both Al and N have oxidation state 3, which simply
AlN(s). It is not soluble in water.

Sodium bromate

Na(BrO₃)₃ (aq)

<u>2. Write the unbalanced molecular equation</u>

Na₃N(aq) + Al(BrO₃)₃(aq) → AlN(s) + Na(BrO₃)₃(aq)

<h2>This is a long answer with more than 5,000 charaters; thus, I have to add the rest of the explanations on a separate file.</h2><h2></h2><h2>The attached file contains the complete answer.</h2>

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