02 (g) + 2H2 (g) If the pressure is increased for this reaction at equilibrium:

Question

02 (g) + 2H2 (g) If the pressure is increased for this reaction at equilibrium: 2H20(g) a. only the K value will change. b. the K value will change and the system will shift to the right. c. the K value will not change and the system will shift to the right. d. the K value will not change and the system will shift to the left. Ka, the equilibrium constant for the dissociation of H20 has a value of 1.8 x 10 16. Similarly, the Ka for the dissociation of methanol (CH3OH) is 6.3 x 10 Based on this information calculate the equilibrium constant for -18 HA + H20 Dissociation Ka equilibrium for any compound HA is If 32.852 g of KMn04 (M. W. 158.0 gmol) is dissolved in water to make 800 mL solution, what is its molarity?

Explanation / Answer

1) answer : option) C.

O2 (g) + 2 H2 (g) <----------> 2 H2O (g)

if pressure increases , then the equilibrium shifts to fewer number of moles of gas. so equilibrium shifts to right side. K value depend only on temperature. so it does not change.

option C) is correct.

2)

CH3OH   +   H2O   ---------------> CH3O-   +   H3O+   Ka = 6.3 x 10^-18

H3O+   +   OH-    ------------>    H2O + H2O           1 / Ka = 5.6 x 10^15

___________________________________________________________

CH3OH + OH-    -------------> CH3O- + H2O          Ka = 3.5 x 10^-2

equilibrium constant = 3.5 x 10^-2

3)

mass of KMnO4 = 32.852 g

moles of KMnO4 = 32.852 / 158.0 = 0.2079 mol

volume = 800 mL = 0.8 L

Molarity = moles / volume

             = 0.2079 / 0.8

Molarity = 0.2599 M

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