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frez [133]
3 years ago
11

Using the clausius clapeyron equation determines the heat of vaporization from experimental measurements of vapor pressure. the

vapor pressure of dichloromethane is measured as a function of temperature and the results are tabulated. determine the heat and vaporization of dichloromethane from the results
Temperature K Vapor pressure (torr)
255 11.3
265 21.0
275 36.8
285 61.5
295 99.0
300 123.8
Chemistry
1 answer:
Temka [501]3 years ago
4 0
The temperature of the pressure is 255
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Given that Kp = 3.5 x 10-4 for the reaction 2 CO(g) <=> C(graphite) + CO2(g), what is the partial pressure of CO2(g) at eq
spin [16.1K]

<u>Answer:</u> The partial pressure of carbon dioxide at equilibrium is 0.0056 atm

<u>Explanation:</u>

The given chemical equation follows:

                     2CO(g)\rightleftharpoons C\text{ (graphite)}+CO_2(g)

<u>Initial:</u>             4.00

<u>At eqllm:</u>       4.00-2x        x                 x

The expression of K_p for above reaction follows:

K_p=\frac{p_{CO_2}}{(p_{CO})^2}

The partial pressure of pure solids and liquids are taken as 1 in the equilibrium constant expression.

We are given:

K_p=3.5\times 10^{-4}

Putting values in above expression, we get:

3.5\times 10^{-4}=\frac{x}{(4-2x)^2}\\\\x=0.0056,718.28

Neglecting the value of x = 718.28 because equilibrium pressure cannot be greater than initial pressure

Partial pressure of CO_2 = 0.0056 atm

Hence, the partial pressure of carbon dioxide at equilibrium is 0.0056 atm

4 0
3 years ago
A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,
Dafna1 [17]

Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

A(aq)+B(aq)\rightleftharpoons C(aq)

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

where,

\Delta G^o = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above relation, we get:

-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k

k=5.45

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of A at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

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The top of the syringe is a circle. You need to compute its area for use in later computations of pressure values. Start by usin
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Explanation:

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What is the molarity of a solution composed of 4.73 g of potassium iodide (KI) dissolved in enough water to make 0.169 L of solu
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Answer:

[KI] = 0.17 M

Explanation:

We determine the moles of solute:

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Molarity (mol/L) is defined as moles of solute in 1L of solution

It is a sort of concentration

M = 0.0285 mol / 0.169L = 0.17 M

Instead of making the division (mol/L), we can also try this rule of three:

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