All categories

2 years ago

For the reaction A (g) → 2 B (g), K = 14.7 at 298 K. What is the value of Q for this reaction at 298 K when ∆G = -20.5 kJ/mol?

Chemistry

1 answer:

zimovet [89]2 years ago

8 0

Answer:

Q= 250= 2,5 x 10^2

Explanation:

ΔG=ΔGº +RTlnQ. R=8,314 J/molK and T=298K

In the balance ΔG=ΔGº +RTlnK=0 → ΔGº= -RTLnK= - 6,7 KJ/mol

ΔG=ΔGº +RTlnQ → -20,5 = -6,7 + 2,5 LnQ

→LnQ= -5,5 → Q=250

You might be interested in

A student prepared an unknown sample by making a dilute solution of the unknown sample. The dilute sample was prepared by adding

telo118 [61]

Answer:

0.27 mM

Explanation:

According to the law of Lambert Beer:

A = C × E × L

Where:

A = Absorbance

C = Concentration

E = molar absorptivity

L = Step Length

If me make C the subject of the formula, we have = A / E × L

We know Absorbance and we can assume L = 1 cm, but being an unknown substance, we must know the molar Absorptivity of it, therefore, only having that value we could calculate the concentration of the mixture using the previous equation.

Then we could use the dilution ratio:

Cc * Vc = Cd * Vd

from the above formula, to fin the concentrated Concentration, we have:

Cc = Cd * Vd / Vc

we then replace the known values.

Vd = Diluted volume 25 mL

Vc = Volume concentrated 5 mL

and we would not know the diluted concentration (Cd), which is what we had to calculate in the first section. Substituting it in the previous equation, we can obtain the initial concentration.

Assuming a molar absorptivity of 5000 M-1 * cm -1 we would have:

Cd = 0.270 / 5000 * 1

= 5.4 x 10 ^ -5 M * (1000 mM / 1 M)

= 0.054 mM

and initial concentration:

Cc = 0.054 mM * (25 mL / 5 mL)

= 0.27 mM.

4 0

2 years ago

A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the equilibrium pressure of NO2 if Kp for the reac

dlinn [17]

Explanation:

Reaction equation for this reaction is as follows.

$NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

It is given that $K_{p}$ = 0.0118.

According to the ICE table,

$NO(g) + SO_{3}(g) \rightarrow NO_{2}(g) + SO_{2}(g)$

Initial: 0.86 0.86 0 0

Change: -x -x +x +x

Equilibrium: 0.86 - x 0.86 - x x x

Hence, value of $K_{p}$ will be calculated as follows.

$K_{p} = \frac{P_{NO_{2}} \times P_{SO_{3}}}{P_{NO} \times P_{SO_{3}}}$

0.0118 = $\frac{x \times x}{(0.86 - x)^{2}}$

x = 0.084 atm

Thus, we can conclude that $P_{NO_{2}}$ is 0.084 atm.

4 0

2 years ago

A flask of volume 2.0 liters, provided with a stopcock, contains oxygen at 20 oC, 1.0 ATM (1.013X105 Pa). The system is heated t

leonid [27]

Answer:

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

2.6592 grams of oxygen remain in the flask.

Explanation:

Volume of the flask remains constant = V = 2.0 L

Initial pressure of the oxygen gas = $P_1=1.0 atm$

Initial temperature of the oxygen gas = $T_1=20^oC =293.15 K$

Final pressure of the oxygen gas = $P_2=?$

Final temperature of the oxygen gas = $T_2=100^oC =373.15 K$

Using Gay Lussac's law:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1\times T_2}{T_1}=\frac{1 atm\times 373.15 K}{293.15 K}=1.27 atm$

1.27 atm is the final pressure of the oxygen in the flask (with the stopcock closed).

Moles of oxygen gas = n

$P_1V_1=nRT_1$ (ideal gas equation)

$n=\frac{P_1V_1}{RT_1}=\frac{1 atm\times 2.0 L}{0.0821 atm l/mol K\times 293.15 K}=0.08310 mol$

Mass of 0.08310 moles of oxygen gas:

0.08310 mol × 32 g/mol = 2.6592 g

2.6592 grams of oxygen remain in the flask.

6 0

2 years ago

The element thallium is 70% thallium-205 and 30% thallium-203. Calculate its relative atomic mass to 1 decimal place.

xxMikexx [17]

Answer:

answer-

The relative atomic mass = 204.4

explanation:

Thallium -203 = 30%

Thallium -205 = 70%

Therefore ,

relative mass of thallium = (30×203 + 70×205)/100

relative mass of thallium = (20440)/100

relative mass of thallium = 204.40 amu

Thus,

relative atomic mass of thalium =204.4 ( to 1 decimal place)

6 0

2 years ago

How many moles of tungsten (W,183.85 g/lol are in 415 grams of tungsten?

vladimir1956 [14]

Given mass of tungsten, W = 415 g

Molar mass of tungsten, W = 183.85 g/mol

Calculating moles of tungsten from mass and molar mass:

$415 g * \frac{1 mol}{183.85 g} = 2.26 mol W$

7 0

2 years ago