Question

n the formation of chloromethane and hydrogen chloride. The overall reaction is 2CH4(g)+3Cl2(g)⟶2CH3Cl(g)+2HCl(g)+2Cl−(g) 2CH4(g)+3Cl2(g)⟶2CH3Cl(g)+2HCl(g)+2Cl−(g) Suppose that a chemist combines 295 mL295 mL of methane and 725 mL725 mL of chlorine at STP in a 2.00 L2.00 L flask. The flask is then allowed to stand at 298 K.298 K. If the reaction reaches 77.0%77.0% completion, what is the total pressure in the flask?

Answer 1

Answer:

The total pressure in the flask is 0,619 atm.

Explanation:

For the reaction:

2CH₄(g) + 3Cl₂(g) ⟶ 2 CH₃Cl(g) + 2HCl(g) + 2Cl⁻(g)

The moles of CH₄ in 295 mL at STP are:

n = PV/RT

Where P is pressure (1 atm), V is volume (0,295L), R is gas constant (0,082atmL/molK) and T is temperature (273,15 K)

Replacing, moles of CH₄ are 0,0132 moles

In the same way, moles of chlorine are 0,0324 moles

As 3 moles of Cl₂ react with 2 moles of CH₄, for a total reaction of 0,0132 moles of CH₄ you need:

0,0132 moles CH₄ × $\frac{3 moles Cl_{2}}{2 moles CH_{4}}$ = 0,0198 moles Cl₂. That means that 0,0324-0,0198 = 0,0126 moles of Cl₂ are in excess.

As the reaction reaches in 77%, the moles of CH₄ that don't react are:

0,0132×(100%-77%)= 3,036x10⁻³ moles of CH₄

Also, the moles of Cl₂ that don't react are:

0,0126 + 0,0198×(100%-77%)= 0,0172 moles of Cl₂

The moles produced of each compound are:

0,0132×77% × $\frac{2 moles CH_{3}Cl}{2 moles CH_{4}}$ = 0,0102 moles of CH₃Cl -that are the same moles of HCl and Cl⁻-

Thus, total moles in the flask are:

3,036x10⁻³ moles of CH₄ + 0,0172 moles of Cl₂ + 0,0102 moles of CH₃Cl + 0,0102 moles of HCl + 0,0102 moles of  Cl⁻ = 0,0507 total moles

As the volume of the flask is 2,00L and the final temperature is 298 K. The total pressure in the flask is:

P = nRT/V

P = 0,619 atm

I hope it helps!