Ask question

Ask question

All categories

soldi70 [24.7K]

3 years ago

6

Consider the following reaction at 25 °C: 4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s) An equilibrium mixture contains 1.0 mol Fe, 1.0 × 10–3 mol

O2, and 2.0 mol of Fe2O3 all in a 2.0 L container. Calculate the value of K for this reaction. What is the value of Kp for this reaction. (Answer: 8.0 × 109 M–3; 5.5 × 105 atm–3)

1 answer:

tino4ka555 [31]3 years ago

4 0

Answer:

Kc = Kc = 8.0 * 10^9

Kp = 5.5 *10^5

Explanation:

Step 1: Data given

Temperature = 25.0 °C

Number of moles Fe = 1.0 moles

Number of moles O2 = 1.0 * 10^-3 moles

Number of moles Fe2O3 = 2.0 moles

Volume = 2.0 L

Step 2: The balanced equation

4Fe(s) + 3O2(g) ⇌ 2Fe2O3(s)

Step 3: Calculate molarity

Molarity = moles / volume

[Fe] = 1.0 moles / 2.0 L

[Fe] = 0.5 M

[O2] = 0.001 moles / 2.0 L

[O2] = 0.0005 M

[Fe2O3] = 2.0 moles / 2.0 L

[Fe2O3] = 1.0 M

Step 4: Calculate Kc

Kc =1/ [O2]³

Kc = 1/0,.000000000125

Kc = 8.0 * 10^9

Step 5: Calculate Kp

Kp = Kc*(R*T)^Δn

⇒with Kc = 8.0*10^9

⇒with R = 0.08206 L*atm /mol*K

⇒with T = 298 K

⇒with Δn = -3

Kp = 8.10^9 *(0.08206 * 298)^-3

Kp = 5.5 *10^5

You might be interested in

Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass

yan [13]

Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

$2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)$

$\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles$

$\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol$

According to stoichiometry :

2 moles of $NaCl$ are formed from = 2 moles of $Na$

Thus 16 moles of $NaCl$ are formed from= $\frac{2}{2}\times 16=16moles$ of $Na$

Mass of $Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g$

Thus 368 grams of sodium reacted.

4 0

3 years ago

What is the percent composition of Br in CuBr3?

tekilochka [14]

Answer:

about 79% (79.04369332 to be exact)

Explanation:

Percent composition=(Molar mass of element x amount of it)/Molar mass of compound x 100

Br= 3 x 79.9/303.25 x100=79.04369332

6 0

3 years ago

Round to 4 significant figures: 42,561

arsen [322]

6 is the fourth significant figures

if the number behind it is 5 or more then 5, you must add 1 to the number and ALL the number behind it will turn into 0

so that the answer is 42560

7 0

3 years ago

Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

What is the total number of liters of NH3 formed when 20 liters of N2 reacts completely

goblinko [34]

Answer:

N2 + H2 ----------》NH3

On balancing it

N2. + 3.H2------->2.NH3

( 1 mol) (3 mol) (2 mol)

1 mol of nitrogen reacts with 3 mol of hydrogen to give 2 mol of ammonia.

Likewise,

20 litres of nitrogen reacts with 60 litres of hydrogen to give 40 litres of Ammonia.

Hence, the answer is 40 Litres.

5 0

3 years ago