Must include answers to all three for points. Answers and work must be shown. If

Question

Must include answers to all three for points. Answers and work must be shown. If your answer is wrong you will not get points and I will choose no best answer.

1) http://i.imgur.com/QQLmrfn.png

2) http://i.imgur.com/2vMjwFo.png

3) http://i.imgur.com/TW59WmW.png

Must include answers to all three for points. Answers and work must be shown. If your answer is wrong you will not get points and I will choose no best answer. http://i.imgur.com/QQLmrfn.png http://i.imgur.com/2vMjwFo.png http://i.imgur.com/TW59WmW.png

Explanation / Answer

1)

H2+O2 gives H2O2 : K1

2H2 + O2 gives 2H2O : K2

H2O + 0.5O2 gives H2O2 : K3

If we reverse the second equation , 2H2O gives 2H2 + O2 : 1/K2

If we multiply this by 0.5, H2O gives H2 + 0.5O2 : (1/K2)^0.5

Adding this with first equation :

H2O + H2 + O2 gives H2+ 0.5O2 + H202 : K1*(1/K2)^0.5

This becomes H2O + 0.5O2 gives H2O2: K1*(1/K2)^0.5

So, K3 = K1*(1/K2)^0.5 = 2.4 * 10^6 * (1/1.8*10^37)^0.5 =( 0.2357*10^-18) * 2.4 * 10^6

= 0.5656 * 10^-12 = 5.66 * 10^-13 = 5.7 * 10^-13 approx . so OPTION-4TH.

2) Kp = Kc (RT)^(n2-n1)

T = 900 C = 900+273 K = 1173K

n2 = moles on right side = 3

n2 = moles on left side = 5

So, n2-n1 = 3-5 = -2.

So, Kp = 0.31 * (0.0821*1173)^-2 = 3.3 * 10^-5

3) Option 2 is correct. When the container is rigid, the system is at constant volume. An inert Gas addition does not affect the position of equilibrium if system is at constant volume. However, when the system has a piston, the volume may change and thus the equilibrimum position shifts. This conclusion is drawn using LeChatelier's principle.

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