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Ratling [72]
2 years ago
15

For the gaseous reaction of carbon monoxide and chlorine to form phosgene (COCl₂):(b) Assuming that ΔS° and ΔH° change little wi

th temperature, calculate ΔG° at 450. K.
Chemistry
1 answer:
maksim [4K]2 years ago
7 0

ΔG° at 450. K is -198.86kJ/mol

The following is the relationship between  ΔG°,  ΔH, and  ΔS°:

ΔH-T ΔS = ΔG

where  ΔG represents the common Gibbs free energy.

the enthalpy change,  ΔH

The temperature in kelvin is T.

Entropy change is  ΔS.

ΔG° = -206 kJ/mol

ΔH° equals -220 kJ/mol

T = 298 K

Using the formula, we obtain:

-220kJ/mol -T ΔS° = -206kJ/mol

220 kJ/mol +206 kJ/mol =T ΔS°.

-T ΔS = 14 kJ/mol

for ΔS-14/298

ΔS=0.047 kJ/mol.K

450K for the temperature Completing a formula with values

ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol

ΔG° = -220 kJ/mol + 21.14 kJ/mol.

ΔG°=198.86 kJ/mol

Learn more about ΔG° here:

brainly.com/question/17214066

#SPJ4

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2 years ago
What electrostatic attraction between____ forms an ionic bond?
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3 years ago
How many grams of calcium phosphate (Ca3(PO4)2) re theoretically produced if we start with 3.40 moles of Ca(NO3)2 and 2.40moles
sattari [20]

1) Balance the chemical equation.

3Ca(NO_3)_2+2Li_3PO_4\rightarrow6LiNO_3+Ca_3(PO_4)_2

2) List the known and unknown quantities.

Reactant 1: Ca(NO3)2.

Amount of substance: 3.40 mol.

Reactant 2: Li3PO4.

Amount of substance: 2.40 mol.

Product: Ca3(PO4)2

Mass: unknown.

3) Which is the limiting reactant?

<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.

<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>

The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.

mol\text{ }Ca(NO_3)_2=2.40\text{ }mol\text{ }Li_3PO_4*\frac{3\text{ }mol\text{ }Ca(NO_3)_2}{2\text{ }mol\text{ }Li_3PO_4}=3.60\text{ }mol\text{ }Ca(NO_3)_2

<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.

4) Moles of Ca3(PO4)2 produced from the limiting reactant.

We have 3.40 mol Ca(NO3)2 of the limiting reactant.

The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

5) Mass of Ca3(PO4)2 produced.

The molar mass of Ca3(PO4)2 is 310.1767 g/mol.

g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.

Option D.

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Answer:

Kp = (Partial pressure H₂O) . (Partial Pressure Cl₂)² / Partial pressure O₂ . (Partial Pressure HCl)⁴

Explanation:

This is the reaction:

4 HCl (g) + O₂ (g)  ⇒ 2 Cl₂ (g)  + 2 H₂O(g)

Kp = (Partial pressure H₂O) . (Partial Pressure Cl₂)² / Partial pressure O₂ . (Partial Pressure HCl)⁴

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