Calculate K for the reactions in Question 2 .

Name: Problem 6, Chapter 14: Chemistry Principles and Reactions
Uploaded: 2021-08-09T00:01:10-08:00
Duration: 5 min 26 s
Channel: Eileen Sullivan
Description: Calculate K for the reactions in Question 2 .

Key Concepts

Gibbs Free Energy and Equilibrium

The link between Gibbs free energy change and the equilibrium constant, expressed by the relation ?G° = -RT ln K, connects thermodynamics with chemical equilibrium. This relationship helps in understanding how the spontaneity of a reaction at a given temperature influences the value of the equilibrium constant.

Equilibrium Constant (K)

The equilibrium constant is a key parameter in chemical reactions, representing the ratio of product concentrations to reactant concentrations, each raised to their respective stoichiometric coefficients, at equilibrium. It provides insight into whether a reaction favors the formation of products or reactants under specific conditions.

Law of Mass Action

This law underpins the derivation of the equilibrium constant expression. It states that the rate at which a reaction proceeds is proportional to the product of the concentrations (or pressures) of the reactants, each raised to the power of its coefficient in the balanced equation, and it helps formalize how K is calculated.

Reaction Stoichiometry

Accurate use of reaction stoichiometry is essential when setting up the equilibrium constant expression. Stoichiometry ensures that the coefficients from the balanced chemical equation correctly dictate the exponents for the concentrations or pressures of the species involved.

Transcript

00:01 So we're going to calculate some ks here.

00:03 Okay, so the first equation that we're talking about here is between acetate and hydrogen ion.

00:15 Please assume that everything is aqueous and less otherwise indicated, and they're going to go ahead and form acetic acid.

00:26 So if you look at this equation, look at the reverse reaction, and you can see that it's the acid ionizing into its ions.

00:34 So the reaction is just the opposite or the reverse of a k -a.

00:39 So the k for this reaction is simply 1 over k -a for acetic acid.

00:50 So we'll go ahead and look that up.

00:52 So that's just going to be 1 over 1 .8 times 10 of the negative 5.

00:57 So the k for this reaction will be 5 .6 times 10 to the 4th.

01:06 Our second reaction is between a strong acid and a strong base.

01:13 So the net ionic equation is simply h plus plus oh minus gives us water.

01:21 And again, if we look at the reverse reaction, we can see it's water ionizing.

01:26 So the k for this reaction is simply one over kw.

01:31 So the reverse reaction would be an ionization of water, and that would be kw.

01:35 The reverse reaction would just be one over.

01:38 So k is going to be one over 1 times 10 of minus 14.

01:43 So we'll get 1 times 10 of the positive 14 for k for this reaction.

01:51 This next one is going to be a little bit more challenging.

01:54 We're going to have to get some reactions to add up to one we're looking for.

01:59 So the one we're looking for here is going to be between hocl and cn minus.

02:09 To make hcn and ocl minus.

02:16 So i'm going to start by writing a reaction for the hydrolysis of our weak acid, h .o .c .l.

02:24 I'm going to go ahead and react that with water.

02:28 So the acid is going to make hydronium ion, and it reacts with water, and leave us with ocl minus.

02:38 Then i'm going to do something similar with cyanide, which is a weak base...

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