Hydrogen-Hydrogen
Oxygen=Oxygen
Answer:
1.169s
Explanation:
k = 0.851 M-1s-1
The unit of the rate constant, k tells us this is a second order reaction.
From the question;
Initial Concentration [A]o = 2.01M
Final Concentration [A] = One third of 2.10 = (1/3) * 2.10 = 0.67M
Time = ?
The integrated rate law for second order reactions is given as;
1 / [A] = (1 / [A]o) + kt
Making t subject of interest, we have;
kt = (1 / [A] ) - (1 / [A]o )
t = (1 / [A] ) - (1 / [A]o ) / k
Inserting the values;
t = [ (1 / 0.67 ) - (1 / 2.10) ] / 0.851
t = ( 1.4925 - 0.4975 ) / 0.851
t = 0.995 / 0.851
t = 1.169s
[A] = 0.13073 M ≈ 0.13 M ( 2 s.f)
Answer: The expression for equilibrium constant is ![\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
Explanation: Equilibrium constant is the expression which relates the concentration of products and reactants preset at equilibrium at constant temperature. It is represented as 
For a general reaction:

The equilibrium constant is written as:
![k_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Chemical reaction for the formation of ammonia is:


Expression for
is:
![k_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=k_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
![1.6\times 10^2=\frac{[NH_3]^2}{[H_2]^3[N_2]}](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E2%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BH_2%5D%5E3%5BN_2%5D%7D)
The enthalpy change for the reaction 2K + S + K2S using the following thermochemical reactions is mathematically given as
dn=70.7kg
<h3>What is the enthalpy change for the reaction?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
2k+s----->K2S
Therefore
KS+K----->K2S
Therefore
dn=dn1+dn2
dn=32.5+38.2
dn=70.7kg
In conclusion, the enthalpy change is
dn=70.7kg
Read more about Chemical reaction
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