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there are two steps in the usual industrial preparation of acrylic acid…

Question

there are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics.

cac₂(s) + 2h₂o(l) ⇌ c₂h₂(g) + ca(oh)₂(s) k₁

6c₂h₂(g) + 3co₂(g) + 4h₂o(l) ⇌ 5ch₂chco₂h(g) k₂

the net reaction is:

6cac₂(s) + 16h₂o(l) + 3co₂(g) ⇌ 6ca(oh)₂(s) + 5ch₂chco₂h(g) k

write an equation that gives the overall equilibrium constant k in terms of the equilibrium constants k₁ and k₂. if you need to include any physical constants, be sure you use their standard symbols, which youll find in the aleks calculator.

k = □

Explanation:

Step1: Manipulate the first reaction

Multiply the first reaction \( \text{CaC}_{2}(s)+2\text{H}_{2}\text{O}(l)
ightleftharpoons\text{C}_{2}\text{H}_{2}(g)+\text{Ca(OH)}_{2}(s) \) by 6.
When a reaction is multiplied by a factor \( n \), its equilibrium constant \( K \) is raised to the power \( n \). So the equilibrium constant for \( 6\text{CaC}_{2}(s)+12\text{H}_{2}\text{O}(l)
ightleftharpoons6\text{C}_{2}\text{H}_{2}(g)+6\text{Ca(OH)}_{2}(s) \) is \( K_{1}^{6} \).

Step2: Combine the manipulated first reaction and the second reaction

The second reaction is \( 6\text{C}_{2}\text{H}_{2}(g)+3\text{CO}_{2}(g)+4\text{H}_{2}\text{O}(l)
ightleftharpoons5\text{CH}_{2}\text{CHCO}_{2}\text{H}(g) \) with equilibrium constant \( K_{2} \).
When two reactions are added together, the equilibrium constant of the net - reaction is the product of the equilibrium constants of the individual reactions.
The net - reaction \( 6\text{CaC}_{2}(s)+16\text{H}_{2}\text{O}(l)+3\text{CO}_{2}(g)
ightleftharpoons6\text{Ca(OH)}_{2}(s)+5\text{CH}_{2}\text{CHCO}_{2}\text{H}(g) \) is obtained by adding \( 6\text{CaC}_{2}(s)+12\text{H}_{2}\text{O}(l)
ightleftharpoons6\text{C}_{2}\text{H}_{2}(g)+6\text{Ca(OH)}_{2}(s) \) and \( 6\text{C}_{2}\text{H}_{2}(g)+3\text{CO}_{2}(g)+4\text{H}_{2}\text{O}(l)
ightleftharpoons5\text{CH}_{2}\text{CHCO}_{2}\text{H}(g) \)

Answer:

\( K = K_{1}^{6}K_{2} \)