4.91 For the reaction $$ \mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{HN}_{3}(g) $$ a. Balance the equation.

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1. Write the unbalanced reaction:  
   $$ \mathrm{N}_2(g) + \mathrm{H}_2(g) \longrightarrow \mathrm{HN}_3(g) $$

2. Assign coefficients to each species:  
   Let the balanced equation be  
   $$
   a\,\mathrm{N}_2(g) + b\,\mathrm{H}_2(g) \longrightarrow c\,\mathrm{HN}_3(g)
   $$

3. Balance the atoms for each element.

- **Nitrogen:**  
     On the left: $$2a$$ atoms (from $$\mathrm{N}_2$$).  
     On the right: $$3c$$ atoms (each $$\mathrm{HN}_3$$ contains 3 nitrogen atoms).  
     Therefore,  
     $$
     2a = 3c
     $$

- **Hydrogen:**  
     On the left: $$2b$$ atoms (from $$\mathrm{H}_2$$).  
     On the right: $$c$$ atoms (each $$\mathrm{HN}_3$$ contains 1 hydrogen atom).  
     Therefore,  
     $$
     2b = c
     $$

4. Choose the smallest integer values to satisfy these equations.

- From the hydrogen balance, we have $$c = 2b$$.  
   - Substitute $$c = 2b$$ into the nitrogen balance:  
     $$
     2a = 3(2b) \quad \Rightarrow \quad 2a = 6b \quad \Rightarrow \quad a = 3b
     $$
   
   - Set $$b = 1$$ (the smallest positive integer):  
     $$
     a = 3 \quad \text{and} \quad c = 2
     $$

5. Write the balanced equation with the coefficients:  
   $$
   3\,\mathrm{N}_2(g) + 1\,\mathrm{H}_2(g) \longrightarrow 2\,\mathrm{HN}_3(g)
   $$

6. Verify the balancing:

- **Nitrogen:**  
     Left: $$3 \times 2 = 6$$ nitrogen atoms.  
     Right: $$2 \times 3 = 6$$ nitrogen atoms.
   
   - **Hydrogen:**  
     Left: $$1 \times 2 = 2$$ hydrogen atoms.  
     Right: $$2 \times 1 = 2$$ hydrogen atoms.

The balanced equation is:  
$$
3\,\mathrm{N}_2(g) + \mathrm{H}_2(g) \longrightarrow 2\,\mathrm{HN}_3(g)
$$
The balanced equation is: $$ 3\,\mathrm{N}_2(g) + \mathrm{H}_2(g) \longrightarrow 2\,\mathrm{HN}_3(g) $$

4.91 For the reaction

a. Balance the equation.

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4.91 For the reaction a. Balance the equation.