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The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) you may want to reference (pages 680 - 684) section 15.3 while completing this problem. part a calculate kp for the reaction below. 4cof2(g)⇌2co2(g)+2cf4(g)

Respuesta :

Kalahira
Kp is the equilibrium pressure constant calculated from the partial pressures of a reaction equation.
 Kp =[pCF4]*[p CO2] / [p COF2]^2 = 2.2 x 10^6
When the mole fraction gets doubled we have
 Kp = [pCO2]^2*[pCF4]^2 / [pCOF2]^4
 Kp = [[pCF4]*[p CO2] / [p COF2]^2] * 2
Kp = (2.2 * 10^6) * 2
Kp = 4.8 * 10^12
dsdrajlin

The equilibrium constant (Kp) for the reaction

4 COF₂(g) ⇌ 2 CO₂(g) + 2 CF₄(g) is 4.8 × 10¹².

Let's consider the following reaction.

2 COF₂(g) ⇌ CO₂(g) + CF₄(g)

Its equilibrium constant is:

[tex]Kp_1 = \frac{[CF_4][CO_2]}{[COF_2]^{2} }[/tex]

Now, let's consider the second reaction.

4 COF₂(g) ⇌ 2 CO₂(g) + 2 CF₄(g)

Its equilibrium constant is:

[tex]Kp_2 = \frac{[CF_4]^{2} [CO_2]^{2}} {[COF_2]^{4} } = \frac{[CF_4] [CO_2]} {[COF_2]^{2} } \times \frac{[CF_4] [CO_2]} {[COF_2]^{2} } = Kp_1 \times Kp_1 = (Kp_1)^{2} = (2.2 \times 10^{6} )^{2} = 4.8 \times 10^{12}[/tex]

The equilibrium constant (Kp) for the reaction

4 COF₂(g) ⇌ 2 CO₂(g) + 2 CF₄(g) is 4.8 × 10¹².

Learn more: https://brainly.com/question/15118952

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