Question

the chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. The overall reaction is : 2CH4(g)+3Cl2(g)⟶2CH3Cl(g)+2HCl(g)+2Cl−(g) Suppose that a chemist combines 295 mL of methane and 725 mL of chlorine at STP in a 2.00 L flask. The flask is then allowed to stand at 298 K. If the reaction reaches 77.0% completion. What is the total pressure in the flask? (in atm) What is the partial pressure of CH4?( in atm)
What is the partial pressure of Cl2?(in atm)
What is the partial pressure of CH3Cl?( in atm)
What is the partial pressure of HCl? ( in atm)
What is the partial pressure of Cl−? ( in atm)

Answer 1

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× $\frac{3molCl_{2}}{2 molCH_{4}}$ = 0,0198 moles

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 = 0,0171 mol Cl₂

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% = 0,0102 moles of CH₄ And the moles that don't react are 0,00300 mol

Thus, moles of each compound are:

0,0102 moles of CH₄×$\frac{3Cl_{2}}{2 molCH_{4}}$= 0,0153 mol  + 0,0171 mol = 0,0324 mol Cl₂

0,0102 moles of CH₄×$\frac{2CH_{3}Cl}{2 molCH_{4}}$= 0,0102 mol CH₄

0,0102 moles of CH₄×$\frac{2HCl}{2 molCH_{4}}$= 0,0102 mol HCl

0,0102 moles of CH₄×$\frac{2Cl^{-}}{2 molCH_{4}}$= 0,0102 mol Cl⁻

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

-To obtain partial pressure you change the moles for each compound-

I hope it helps!

Answer 2

The total pressure in the 2 L flask is 0.806 atm. The total pressure of the reaction can be calculated by the ideal gas equation.

How to calculate the pressure?

The total pressure of the reaction can be calculated by the ideal gas equation,

$P = \dfrac {nRT}V$

Where,

$P$ - pressure

$V$ - volume =          2 L

$R$ - gas constant =  0.082 atmL/molK

$T$ - temperature =   298 K

$n$ - number of moles = 0.066 mol

Put the values in the formula,

$P = \dfrac {0.066 \times 0.082 \times 298} 2 \\\\P = 0.806\rm \ atm$

Therefore, the total pressure in the 2 L flask is 0.806 atm.

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