Write the balanced chemical equation for the decomposition of sodium azide. If this is a redox reaction, show half reactions and determine if spontaneous. The chemical equation is 2NaN3(s) -> 2Na(s)+3N2(g) right? I am confused on the oxidation states for NaN3... please explain!

The balanced chemical equation for the decomposition of sodium azide (NaN3) is:

2NaN3(s) -> 2Na(s) + 3N2(g)

This equation indicates that when solid sodium azide decomposes, it produces solid sodium and gaseous nitrogen.

To determine the oxidation states for NaN3 (sodium azide), we need to consider the oxidation states of individual elements in the compound.

In sodium azide (NaN3), sodium (Na) is an alkali metal and typically has an oxidation state of +1. The three nitrogen atoms are all identical, so they will have the same oxidation state.

To determine the oxidation state of nitrogen, we assign the variable x to it. Since there are three nitrogen atoms in NaN3, the sum of all oxidation states should be zero (because the compound is neutral):

(+1) + 3x = 0

Solving the equation, we find that x = -1/3.

Therefore, the oxidation state of each nitrogen atom in sodium azide is -1/3. This is an unusual oxidation state, but it is a result of being stabilized in the compound NaN3.

To determine if this decomposition reaction is a redox reaction and if it is spontaneous, let's examine the oxidation states.

In NaN3, the oxidation state of sodium changes from +1 to 0, indicating that sodium is reduced. The oxidation state of nitrogen changes from -1/3 to 0, indicating that nitrogen is oxidized.

The sodium reduction half-reaction is: Na+ + e- -> Na (reduced form)
The nitrogen oxidation half-reaction is: N(-1/3) -> N(0) + (1/3)e- (oxidized form)

Since there is a change in oxidation states for both sodium and nitrogen, this decomposition of sodium azide is indeed a redox reaction.

Determining spontaneity requires considering the standard reduction potentials of the species involved. Without the values of the reduction potentials, it is difficult to determine spontaneity conclusively.

To balance the chemical equation for the decomposition of sodium azide, we start by writing the formula for the reactant and the products.

Sodium azide can be represented by the formula NaN3. It decomposes into sodium (Na) and nitrogen gas (N2). The balanced chemical equation for this reaction is:

2 NaN3(s) -> 2 Na(s) + 3 N2(g)

Now let's determine the oxidation states for each element involved in the reaction:

In sodium azide (NaN3), oxygen (O) is usually assigned an oxidation state of -2. Since there are no oxygen atoms in NaN3, we don't need to consider it.

The general guideline for assigning oxidation states is as follows:
- The oxidation state of an element in its elemental form is always 0.
- The oxidation state of a monatomic ion is equal to its charge.

To determine the oxidation state of sodium (Na), we consider that it is a Group 1A element, meaning it tends to lose one electron to attain a noble gas configuration. Therefore, the oxidation state of sodium is +1.

In the case of nitrogen (N), it usually has an oxidation state of -3 in compounds (except in certain cases like in ammonia, where it is -3). So, in sodium azide (NaN3), since there are three nitrogen atoms, we can assign an oxidation state of -3 to each nitrogen atom.

By applying these oxidation states, we can write the half-reactions for the redox reaction:

Oxidation half-reaction:
NaN3(s) -> Na(s) + N3(-3) + 3e-

Reduction half-reaction:
N3(-3) + 3e- -> 3N2(g)

Now, let's check if it is a spontaneous reaction. We can do this by comparing the reduction potentials of the half-reactions. However, since the standard reduction potentials for these half-reactions are not provided, we cannot determine the spontaneity of the reaction based on this information alone.

Note: It is important to exercise caution when dealing with sodium azide, as it is a highly toxic and potentially explosive compound.

On the left side,

Na is +1
Each N in NaN3 is -1/3

On the right side, Na is zero, N in N2 is zero.