What is the coefficient of h2 when you balance the equation for this redox reaction? no2(g) + h2(g) → nh3(g) + h2o(l)?

To balance any redox equation, first divide the reaction into two half reactions, and balance the atom whose oxidation number is changing: 

NO2 --> NH3 
H2 --> H2O 

Next, balance oxygens by adding H2O where needed, and then balance hydrogen by adding H+: 

NO2 --> NH3 + 2 H2O 
H2O + H2 --> H2O 

7 H+ + NO2 --> NH3 + 2 H2O 
H2O + H2 --> H2O + 2 H+ 

Next, balance charges by adding electrons (e-): 

7 H+ + NO2 + 7 e- --> NH3 + 2 H2O 
H2O + H2 --> H2O + 2 H+ + 2 e- 

Now, multiply one or both equations by small numbers to make the number of electrons the same in the two half reactions. In this case, multiply the top equation by 2 and the bottom one by 7 to give you 14 electrons in each half reaction: 

14 H+ + 2 NO2 + 14 e- --> 2 NH3 + 4 H2O 
7 H2O + 7 H2 --> 7 H2O + 14 H+ + 14 e- 

Finally, add the two half reactions and simplify by cancelling anything that is on both sides. This will give you the final balanced reaction: 

2 NO2 + 7 H2 --> 2 NH3 + 4 H2O 

So, the answer to the original question is (A) 7

Ea = 42,0 kJ/mol

Explanation:

It is possible to solve this question using Arrhenius formula:

Where:

k1: 3,36x10⁴ M⁻¹ s⁻¹

T1: 344K

Ea = ???

R = 8,314472x10⁻³ kJ/molK

k2 : 7,69 M ⁻¹ s⁻¹

T2: 219K

Solving:

Ea = 42,0 kJ/mol

I hope it helps!