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일반화학/[15장] 화학 평형

1.00 L 0.298 mol PCl3(g) 8.70×10^(-3) mol PCl5(g) 2.00×10^(-3) mol Cl2(g)

by 영원파란 2021. 6. 10.

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1.00 L 0.298 mol PCl3(g) 8.70×10^(-3) mol PCl5(g) 2.00×10^(-3) mol Cl2(g)

 

 

At a certain temperature a 1.00 L flask initially contained

0.298 mol PCl3(g) and 8.70×10^(-3) mol PCl5(g).

After the system had reached equilibrium,

2.00×10^(-3) mol Cl2(g) was found in the flask.

Gaseous PCl5 decomposes according to the reaction

PCl5(g) ⇌ PCl3(g) + Cl2(g)

Calculate the equilibrium concentrations of all species

and the value of K.

 

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ICE 도표를 작성하면,

 

 

 

 

평형에서, x = 0.00200 mol

 

> PCl5의 몰수 = 0.00870 – 0.00200 = 0.00670 mol

> PCl3의 몰수 = 0.298 + 0.00200 = 0.300 mol

> Cl2의 몰수 = 0.00200 mol

 

 

 

플라스크의 부피 = 1.00 L 이므로,

각 성분의 몰수 = 몰농도

 

 

 

K = [PCl3] [Cl2] / [PCl5]

= (0.300) (0.00200) / (0.00670)

= 0.089552...

= 0.0896

 

 

 

 

[키워드] 1.00 L 0.298 mol PCl3(g) 8.70×10^(-3) mol PCl5(g) PCl5(g) ⇌ PCl3(g) + Cl2(g) [Cl2] = 2.00×10^(-3) mol

 

 

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