평형상수. NH3 0.0150 mol O2 0.0150 mol 1.00 L

평형상수. NH3 0.0150 mol O2 0.0150 mol 1.00 L

 

 

다음 반응을 생각하자.

4NH3(g) + 3O2(g) ⇌ 2N2(g) + 6H2O(g)

암모니아 0.0150 mol, 산소 0.0150 mol을 어떤 온도에서 1.00 L 용기에 넣었을 때 평형에서 질소의 농도는 1.96*10^-3 M 이었다. 이 반응의 평형상수 Kc를 구하시오.

 

 

The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:

4NH3(g) + 3O2(g) ⇌ 2N2(g) + 6H2O(g).

When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 1.96*10^-3 M. Calculate Kc.

 

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ICE 도표를 작성하면,

 

	4NH3	+	3O2	⇌	2N2	+	6H2O
초기(I):	0.0150		0.0150		0		0
변화(C):	-4x		-3x		+2x		+6x
평형(E):	0.0150-4x		0.0150-3x		2x		6x

 

 

2x = 1.96*10^(-3) M

x = (1.96*10^(-3)) / 2 = 0.00098 M

 

 

Kc = [N2]^2 [H2O]^6 / [NH3]^4 [O2]^3

= [(1.96*10^(-3))^2 (6*0.00098)^6]

/ [(0.0150-4*0.00098)^4 (0.0150-3*0.00098)^3]

= 6.0059*10^(-6)

 

 

답: 6.01*10^(-6)