Sometimes called the "River of Grass," ________

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

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

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

2. What is the concentration of hydrochloric acid?

nika2105 [10]

<h2>Answer :-</h2><h3>Hydrochloric acid (HCI) is supplied commerically at concentrations mainly within the range 28-36% w/w.</h3>

<h2>Please mark may be brainliest answer. </h2>

3 0

2 years ago

If 0.092J of heat causes a 0.267 degree C temperature change, what mass of water is present?

Thepotemich [5.8K]

Answer:

0.082g

Explanation:

The following data were obtained from the question:

Heat (Q) = 0.092J

Change in temperature (ΔT) = 0.267°C

Specific heat capacity (C) of water = 4.184J/g°C

Mass (M) =..?

Thus, the mass of present can be obtained as follow:

Q = MCΔT

0.092 = M x 4.184 x 0.267

Divide both side by 4.184 x 0.267

M = 0.092 / (4.184 x 0.267)

M = 0.082g

Therefore, mass of water was present is 0.082.

6 0

3 years ago

Which is the fastest in the universe apart from the light

Salsk061 [2.6K]

Well i did some research and i found out now im not sure if this is completely true but supposedly tachyons are even faster than the speed of light

3 0

3 years ago

Radon (Rn) is the heaviest, and only radioactive, member of Group 8A(18) (noble gases). It is a product of the disintegration of

Serggg [28]

Given data Atomic mass of Ra= 226g/mol

no. of moles =1.0/226g/mol =0.04424moles

no. of atoms in 0.044moles

no. of atoms =no. of moles x avogadro's number

= 0.044x 6.022 x10^23 = 0.264968 x 10^22

If 10^15 atoms of Ra produce 1,373*10^4 atoms of Rn per second then 2,66 *10^21 forms 3,658*10^10 atoms of Rn per second.

Day has 246060=86400 s

That means that 2,66x10^21 atoms of Ra produces 3,16 x10^15 atoms of Rn in a day.

N(Rn)=3.16* 10 ^15 n(Rn)=N/NA

n(Rn)=5,25*10−9 pV=nR*T

T=273.15K R=8,314

p=101325Pa V=n∗R∗T/p

V=5.25∗10^−9 ∗ 8.314 ∗ 273.15 / 101325

V=1.1810^−10 m^3 = 118 x10^-7 liters of Rn, measured at STP, are produced per day by 1.0 g of Ra

To know more about Ra here

brainly.com/question/9112754

#SPJ4

6 0

2 years ago

Sometimes called the "River of Grass," _________ Is a vast wetland