The coordination number of cation and anion in fluorite CaF2 and antifluorite Na2O are respectively
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Answer:
Total pressure = 0,806 atm
Partial pressure of CH₄: 0,037 atm
Partial pressure of Cl₂: 0,396 atm
Partial pressure of CH₃Cl: 0,125 atm
Partial pressure of HCl: 0,125 atm
Partial pressure of Cl⁻: 0,125 atm
Explanation:
For the reaction:
2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)
295 mL≡ 0,295L of methane at STP are:
n = PV/RT
P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.
moles of methane: 0,0132 moles
For 725 mL of chlorine ≡ 0,725L
moles of chlorine at STP are: ≡ 0,0324 moles
For a complete reaction of 0,0132 moles of CH₄:
0,0132 mol CH₄× = <em>0,0198 moles</em>
The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol
As you have 0,0324 moles of Cl₂, moles that will not react are:
0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>
As the reaction reaches 77% completion, moles of CH₄ that react are:
0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>
Thus, moles of each compound are:
0,0102 moles of CH₄×= <em>0,0153 mol + 0,0171 mol = 0,0324 mol Cl₂</em>
0,0102 moles of CH₄×= <em>0,0102 mol CH₄</em>
0,0102 moles of CH₄×= <em>0,0102 mol HCl</em>
0,0102 moles of CH₄×= <em>0,0102 mol Cl⁻</em>
Total pressure using:
P = nRT/V
Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L
Total pressure = 0,806 atm
Partial pressure of CH₄: 0,037 atm
Partial pressure of Cl₂: 0,396 atm
Partial pressure of CH₃Cl: 0,125 atm
Partial pressure of HCl: 0,125 atm
Partial pressure of Cl⁻: 0,125 atm
<em>-To obtain partial pressure you change the moles for each compound-</em>
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I hope it helps!
The equilibrium constant is 0.0022.
Explanation:
The values given in the problem is
ΔG° = 1.22 ×10⁵ J/mol
T = 2400 K.
R = 8.314 J mol⁻¹ K⁻¹
The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.
The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K
So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.
We get,
So, the equilibrium constant is 0.0022.