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The reaction 2CH4(g)⇌C2H2(g)+3H2(g) has an equilibrium constant of K = 0.154. If 6.70 mol of CH4, 4.05 mol of C2H2, and 10.05 mol of H2 are added to a reaction vessel with a volume of 5.20 L , what net reaction will occur?

Respuesta :

Answer:

Q>> K

the reaction will proceed to the left

Explanation:

Step 1: Data given

K = 0.154

Number moles CH4 = 6.70 moles

Number of moles C2H2 = 4.05 moles

Number of moles H2 = 10.05 moles

Volume = 5.20 L

Step 2: Calculate concentration of CH4

Concentration = moles / volume

[CH4] = 6.70 mol / 5.20 L = 1.29 M

Step 3: Calculate concentration C2H2

[C2H2] = 4.05 mol / 5.20 L = 0.779 M

Step 3: Calculate concentration H2

[H2] = 10.05 mol / 5.20 L = 1.93 M

Q = [C2H2] [H2]³ / [CH4]²

Q = 0.779 ( 1.93)³ / ( 1.29)² = 3.37 >> K

Q>> K

the reaction will proceed to the left

pstnonsonjoku

Q > K hence the reaction proceeds in the reverse direction and the equation is; C2H2(g)+3H2(g) )⇌ 2CH4(g).

We must first obtain the concentration of each specie as follows;

  • For CH4; 6.70 mol/ 5.20 L = 1.29 M
  • For C2H2; 4.05 mol/ 5.20 L = 0.78 M
  • For H2; 10.05 mol/5.20 L = 1.93 M

Now we have to obtain the reaction quotient Q

Q = [0.78 M] [1.93 M]^3 / [1.29 M]^2

Q = 5.6/1.7

Q= 3.3

We can see from the calculation that Q > K hence the reaction proceeds in the reverse direction and the equation is; C2H2(g)+3H2(g) )⇌ 2CH4(g).

Learn more about equilibrium constant: https://brainly.com/question/17960050

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