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3 years ago

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For a certain chemical reaction, the standard Gibbs free energy of reaction at 10.0 °C is 149. kJ. Calculate the equilibrium con

stant K for this reaction to 2 significant digits. Round your answer K ?

2 answers:

leonid [27]3 years ago

8 0

Answer:

This is confusing

Explanation:

marishachu [46]3 years ago

7 0

<u>Answer:</u> The equilibrium constant for this reaction is $3.1\times 10^{-28}$

<u>Explanation:</u>

To calculate the equilibrium constant (at 15°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = 149. kJ/mol = 149000 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = $8.314J/K mol$

T = temperature = $15^oC=[273+15]K=283K$

$K_{eq}$ = equilibrium constant at 10°C = ?

Putting values in above equation, we get:

$149000J/mol=-(8.314J/Kmol)\times 283K\times \ln K_{eq}\\\\K_{eq}=3.1\times 10^{-28}$

Hence, the equilibrium constant for this reaction is $3.1\times 10^{-28}$

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A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

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Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow

andreyandreev [35.5K]

Explanation:

(a) The given data is as follows.

Pressure on top ( $P_{o}$ ) = 140 bar = $1.4 \times 10^{7} Pa$ (as 1 bar = $10^{5}$ )

Temperature = $15^{o}C$ = (15 + 273) K = 288 K

Density of gas = $\frac{PM}{ZRT}$

$\frac{dP}{dZ} = \rho \times g$

$\frac{dP}{dZ} = \int \frac{PM}{ZRT}$

$\int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ$

$ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}$

= 0.4548

$P_{1} = P_{o} \times e^{0.4548}$

= $1.4 \times 10^{7} Pa \times 1.5797$

= $2.206 \times 10^{7} Pa$

Hence, pressure at the natural gas-oil interface is $2.206 \times 10^{7} Pa$ .

(b) At the bottom of the tank,

$P_{2} = P_{1} + \rho \times g \times h$

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

= $309.8 \times 10^{5} Pa$

= 309.8 bar

Hence, at the bottom of the well at $15^{o}C$ pressure is 309.8 bar.

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3 years ago

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mylen [45]

Here’s the math for your answer, which is 3.3 L HCl

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Anna [14]

Answer: a. The concentrations of the reactants and products have reached constant values

Explanation:

The reactions which do not go on completion and in which the reactant forms product and the products goes back to the reactants simultaneously are known as equilibrium reactions. For a chemical equilibrium reaction, equilibrium state is achieved when the rate of forward reaction becomes equal to rate of the backward reaction.

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time and are constant.

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For a equilibrium reaction,

$A\rightleftharpoons B$

$K_{eq}=\frac{[B]}{[A]}$

Thus the correct answer is the concentrations of the reactants and products have reached constant values.

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3 years ago