Consider a chemical reactionCO(g) + 2H_2(g) = CH_3OH(g). When 2 moles of CO were

Question

Consider a chemical reactionCO(g) + 2H_2(g) = CH_3OH(g). When 2 moles of CO were reacting with 4 moles of H_2 at temperature 500K and pressure p = 100 bar, and chemical equilibrium was reached, n = 0.366 moles of CO was consumed. Based on the given data determine the value of the chemical equilibrium constant K(T) for T = 500. Consider now the same reaction, when 3 moles of CO are reacting with 3 moles of H_2 at the temperature T = 500K and the pressure p_0 = 1 bar. Find the mass w of the product, i.e., CH_3OH, obtained in the reaction.

Explanation / Answer

a) Given that 2 moles of CO reacting wih 4 moles of H2.

CO(g) + 2H2 (g) <---------> CH3OH (g)

Initial 2 mol 4 mol

At equilibrium 2-X 4-2X X

Given that when equilibrium was reached 0.366 moles of CO was consumed.

That means X = 0.366

Therefore, equilibrium concentrations are

[CO] = 2-X = 2-0.366 = 1.634 M

[H2] = 4-2X = 4 - 2 x 0.366 = 3.268 M

[CH3OH] = X = 3.268 M

Hence,

K = [CH3OH]/[CO][H2]2

   = (0.366) / (1.634) (3.268)2

      = 0.021

K = 0.021

Therefore, thermodynmic equilibrim constant K = 0.021

b)

Given that 3 moles of CO reacting wih 3 moles of H2.

CO(g) + 2H2 (g) <---------> CH3OH (g)

From this balanced equation , we can say that

For every 2 moles of H2 , 1 mole of CO is required.

Hence,

For every 3 moles of H2 , 1.5 moles of CO is required.

But we have 3 moles of CO

Therefore, CO is in excess.

Hence, H2 is the limiting reagent.

Therefore, mass of CH3OH is calculated based upon H2.

  CO(g) + 2H2 (g) <---------> CH3OH (g)

2 mol 1 mol= 32 g

3 mol ?

? = (3 mol H2/ 2 mol H2) X 32 g of CH3OH

= 48 g of CH3OH

Therefore, mass of CH3OH obtained in the reaction = 48 g

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.