Dalam ruang 1 liter dicampur 4 mol gas A
dan 5 mo...

Question

Dalam ruang 1 liter dicampur 4 mol gas A 
dan 5 mol gas B. Reaksi kesetimbangan yang 
terjadi adalah: · 
A(g) + 2B(g) ⇌ C(g)
Jika pada keadaan setimbang diperoleh 2 
mol gas C dan tekanan total 10 atm, maka 
harga K adalah .... p 
A. 0,80 
B. 0,60 
c. 0,50 
D. 0,25 
E. 0,05

Kevin L · Answer

The given reaction is A(g) + 2B(g)  C(g). Initially, there are 4 mol of gas A and 5 mol of gas B in a 1 liter container. At equilibrium, 2 mol of gas C is formed.

From the balanced equation, the stoichiometry is 1:2:1 for A:B:C. This means that for every 1 mol of A that reacts, 2 mol of B react and 1 mol of C is produced.

Let x be the number of moles of A that react. Since 2 mol of C is formed, 2x moles of B react. Therefore, the equilibrium concentrations are: [A] = 4 - x, [B] = 5 - 2x, and [C] = 2x.

Given that the total pressure at equilibrium is 10 atm, we can use the ideal gas law to relate the partial pressures to the number of moles and the total pressure. The partial pressures are: PA = (4 - x)RT, PB = (5 - 2x)RT, and PC = 2xRT.

The equilibrium constant Kp is given by: Kp = PC/PA . PB². Substituting the expressions for the partial pressures, we get: 
Kp = 2xRT/(4 - x)RT.(5-2x)RT² 
      = 2x/(4 - x)(5-2x)

Given that PC = 2 mol and total pressure P = 10 atm, we can find x and then calculate Kp. Solving for x, we get x = 1. Substituting x = 1 into the expression for Kp, we find Kp = 0.5. Therefore, the correct answer is C.

Dalam ruang 1 liter dicampur 4 mol gas A dan 5 mol gas B. Reaksi kesetimbangan yang terjadi adalah: · A(g) + 2B(g) ⇌ C(g) Jika pada keadaan setimbang diperoleh 2 mol gas C dan tekanan total 10 atm, maka harga K adalah .... p A. 0,80 B. 0,60 c. 0,50 D. 0,25 E. 0,05

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