The equilibrium constant, Kp, equals 3.40 at 25°C for the isomerization reaction: cis-2-butene ⇌ trans-2-butene. If a flask initially contains 5.00 atm of each gas, the system is___________.

Answers

Answer 1

Answer:

Final answer:

The given reaction will shift towards cis-2-butene once placed in equilibrium. This can be determined by calculating the reaction quotient and comparing it with the equilibrium constant.

Explanation:

The reaction could either shift towards the cis-2-butene or trans-2-butene depending on whether the reaction quotient, Q, is lesser or greater than the equilibrium constant, Kp.

Bear in mind that Kp = Ptrans/Pcis. Let's say that Pt is the partial pressure of trans-2-butene and Pc is the partial pressure of cis-2-butene at equilibrium. If we start with 5 atm of each gas, the change in Pc is -x and the change in Pt is +x.

So, Kp = (5+x)/(5-x). We are given that Kp = 3.4. Solving these two equations will show that x is a negative value, which means that the system shifts towards cis-2-butene.

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Answer 2

Answer:

Final answer:

For the isomerization reaction cis-2-butene ⇌ trans-2-butene, with an initial pressure of 5.00 atm for both gases and a Kp of 3.40, the system will shift towards the product, trans-2-butene, as Kp > Qp (1). This reflects the principle that a chemical system at equilibrium will shift to counteract any change.

Explanation:

In terms of the equilibrium constant (K), for gas-phase reactions, Kp represents equilibrium in terms of partial pressures, while Kc represents it in molar concentrations. For instance, in the isomerization reaction given cis-2-butene ⇌ trans-2-butene, Kp is given as 3.40. To determine the behavior of the system, we need to compare it to reaction quotient (Q). Given that the flask initially contains 5.00 atm of each gas, Qp is 1 (since Qp = partial pressure of trans-2-butene / partial pressure of cis-2-butene). Since Kp > Qp, the reaction will shift towards the products, hence the system will shift towards trans-2-butene. From this, it is clear that the equilibrium constant and reaction quotient play vital roles in determining the direction of shift in a chemical equilibrium.

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4.289x10^0 as regular numbers

Answers

4.289

Explanation:

4.289 x 10^0 can be written as a regular number by moving the decimal point to the right or left based on the exponent value. Since the exponent is 0, the decimal point does not need to be moved. Therefore, the regular number form of 4.289 x 10^0 is simply 4.289.

The heat capacity of chloroform (trichloromethane,CHCl3)in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.

Answers

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

$\Delta S=n\int\limits^(T_f)_(T_i){(C_(p,m)dT)/(T)$

where,

$\Delta S$ = change in molar entropy

n = number of moles = 1.0 mol

$T_f$ = final temperature = 300 K

$T_i$ = initial temperature = 273 K

$C_(p,m)$ = heat capacity of chloroform = $91.47+7.5* 10^(-2)(T/K)$

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

$\Delta S=1.0\int\limits^(300)_(273){((91.47+7.5* 10^(-2)(T/K))dT)/(T)$

$\Delta S=1.0* [91.47\ln T+7.5* 10^(-2)T]^(300)_(273)$

$\Delta S=1.0* 91.47\ln ((T_f)/(T_i))+7.5* 10^(-2)(T_f-T_i)$

$\Delta S=1.0* 91.47\ln ((300)/(273))+7.5* 10^(-2)(300-273)$

$\Delta S=8.626+2.025$

$\Delta S=10.651J/K.mol$

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

the ionization constant for dichloroacetic acid HC2HO2Cl2 is 5.0 x 10^-2 . What is the pH of a 0.15 molar solution of this acid?

Answers

Answer: pH = 1.19

Explanation:

The formula for Ka is: Ka = [H+][A-]/[HA]

where: [H+] = concentration of H+ ions

[A-] = concentration of conjugate base ions

[HA] = concentration of undissociated acid molecules

Equation of reaction: Cl₂CHCOOH ---> H+ + Cl₂CHCOO-

From the equation above, dichloroacetic acid dissociates one H+ ion for every Cl₂CHCOO- ion,

so [H+] = [Cl₂CHCOO-].

Let x represent the concentration of H+ that dissociates from HA, then [HA] = C - x where C is the initial concentration.

Substituting these values into the Ka equation:

Ka = x · x / (C -x)

Ka = x²/(C - x)

(C - x)Ka = x²

x² = CKa - Kax

x² + Kax - CKa = 0

Solve for x using the quadratic formula:

x = [-b ± √(b² - 4ac)]/2a

Note: There are two solutions for x. However only the positive value of x is used since x represents a concentration of ions in solution, and so cannot be negative.

x = [-Ka + √(Ka² + 4CKa)]/2

Substitute the values for Ka and C in the equation above:

Ka = 5.0 x 10^-2

C = 0.15 M

x = {-5.0 x 10^-2 + √[(5.0. x 10^-2)² + 4(0.15)(5.0 x 10^-2)]}/2

x = (-5.0 x 10^-2 + 1.80 x 10^-1)/2

x = 0.13/2

x = 6.50 x 10^-2

To find pH, we use the formula;

pH = -log[H+]

pH = -log(x)

pH = -log(6.50 x 10^-2)

pH = -(-1.19)

pH = 1.19

Two containers, one with a volume of 3.0 L and the other with a volume of 2.0 L contain, respectively, argon gas at 1.1 atm and helium at 0.75 atm. The containers are initially separated by a valve, and then the valve is opened to connect the two containers. Assume perfect gases and determine the followings.a. The total pressure of the mixed gases
b. The partial pressure of each gas
c. The mole fraction of each gas

Answers

Answer:

a. $p_T=0.93atm$ .

$p_(Ar)=0.66atm\n\np_(He)=0.3atm$

$x_(Ar)=0.6875\n\nx_(He)=0.3125$

Explanation:

Hello,

In this case, considering that the valve is opened, we can use the Boyle's law in order to compute the final pressure of argon by considering its initial pressure and volume and a final volume of 5.0 L:

$p_(Ar)=(1.1atm*3.0L)/(5.0L)=0.66atm$

And the final pressure of helium:

$p_(He)=(0.75atm*2.0L)/(5.0L)=0.3atm$

Which actually are the partial pressure of both of them, it means that the total pressure is:

Finally, the mole fraction of each gas is computed by considering the Dalton's law:

$x_i=(p_i)/(p_T)$

$x_(Ar)=(0.66atm)/(0.93atm) =0.6875\n\nx_(He)=(0.3atm)/(0.93atm) =0.3125$

Best regards.

4Ga + 3S2 → 2Ga2S3 1. How many grams of Gallium Sulfide would form if 20.5 moles of Gallium burned?

Answers

Answer:

$m_(Ga_2S_3)=2415.31gGa_2S_3$

Explanation:

Hello,

In this case, for the given chemical reaction, we can notice there is a 4:2 molar ratio between the burned moles of gallium and the yielded moles of gallium sulfide, therefore, we compute them as shown below:

$n_(Ga_2S_3)=20.5molGa*(2molGa_2S_3)/(4molGa)=10.25mol Ga_2S_3$

Then, by using the molar mass of gallium sulfide (235.64 g/mol), we directly compute the grams:

$m_(Ga_2S_3)=10.25mol Ga_2S_3*(235.64gGa_2S_3)/(1molGa_2S_3) \n\nm_(Ga_2S_3)=2415.31gGa_2S_3$

Best regards.

If a cell is 80% water and the outside environment is 90% water. What is likely to happen? A. water will rush into the cell
B. net movement of water will be equal
C. water will not move into or out of the cell
D. water will rush out of the cell

Answers

Water will rush into the cell, because water likes everything to be equal. This process is called diffusion (osmosis).

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