The value of the equilibrium constant for the following chemical reaction (the auto-ionization of water) is 1.0x10-14 at 298 K.
2H2O(l) ßà OH-(aq) + H3O+(aq) K = 1.0x10-14
Using this information and your equilibrium identities, select the correct value for the equilibrium expression for the following chemical reaction (at 298 K as well):
6H2O(l) ßà 3OH-(aq) + 3H3O+(aq) K = ?
Question 5 options:
1x1014 |
|
1x10-28 |
|
1x10-14 |
|
1x10-42 |
|
3x10-14 |
|
1x10-15 |
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The value of the equilibrium constant for the following chemical reaction (the auto-ionization of water) is...
The value for the equilibrium constant for the following chemical reaction, the auto-ionization of water, is 1.0x10-14 at 298 K 2H2O(l) ßà OH-(aq) + H3O+(aq) K = 1.0x10-14 Using this information and your equilibrium identities, select the correct value for the equilibrium expression for the reaction shown below (at 298 K as well) : OH-(aq) + H3O+(aq) ßà 2H2O(l) K = ? Question 4 options: 1x10-14 0.5x10-14 2x10-14 -1x10-14 1x1014 1x10-15
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Use the two given equations to determine the K3 value for the desired equilibrium equation. Given 1: 2H2O(l)= H3O+(aq)+OH-(aq) K1=1.00X10^-14 Given 2: HCN(aq)+H2O(l)=H3O+(aq)+CN-(aq) K2=4.90X10^-10 Given 3: CN-(aq)+H2O(l)=HCN(aq)+OH-(aq) K3=??? What is the K3 value?
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