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Anit [1.1K]
3 years ago
9

For the reaction of gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia, the correct equilibrium constant expressi

on is _____
Chemistry
1 answer:
exis [7]3 years ago
7 0

The solution would be like this for this specific problem:

 

First, we need to write out the balanced reaction equation for the problem:<span>

"gaseous nitrogen with gaseous hydrogen to produce gaseous ammonia" 

gaseous nitrogen + gaseous hydrogen = gaseous ammonia 

gaseous nitrogen = N2(g) 
gaseous hydrogen = H2(g) 
gaseous ammonia = NH3(g) 

In here, I’m going to show you how to balance this from start to finish: 

N2(g) + H2(g) ↔ NH3(g) (basic, unbalanced equation that shows reactants and product) 
N2(g) + H2(g) ↔ 2NH3(g) (balances the nitrogens) 
N2(g) + 3H2(g) ↔ 2NH3(g) (balances the hydrogens) => Balanced Equation 

<span>Next, we write the Keq expression of the balanced reaction:

Keq = (∏[products]^n) /(∏[reactants]^n) 
Keq = [NH3]^2 / [[N2][H2]]^3 


Therefore, the correct equilibrium constant expression is Keq = (NH3)2 / (N2)(H2)3.</span></span>

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Answer:

mole fraction of N_2 O = 0.330

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Explanation:

Given data:

volume of tank 8 L

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Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

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mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

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PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

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Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

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