For the reaction N2 + 3H2 ⇋ 2NH3, the value of KP is 3.6×10^-2at 500K calculate the value of KC for the reaction at the same temperature [R=0.0831]
Answers
Answer:
Initial mole ratio N2:H2 is 1:3
At equilibrium, 50% of N2, H2 has reacted and equilibrium pressure is P
N
2
+3H
2
→2NH
3
y3y2y
At equilibrium, 50% of each reactant reacted, the number of moles will become half.
The mole of N2=y−
2
y
The moles of H2=3y−
2
3y
THE moles of NH3=
2
2y
Total moles at equilibrium=
2
y
+
2
3y
+
2
2y
=3y
Partial pressure of H2=
Totalmoles
MolesofH2
×Totalpressure=
2
3y
×
3y
1
×P=
2
P
Answer:
Value of \sf{K_c}K
c
is \sf{8.66\times\:10^{-4}}8.66×10
−4
.
Explanation:
Given:
Reaction : \sf{N_2+3H_2\leftrightharpoons\:2NH_3}N
2
+3H
2
⇋2NH
3
\sf{K_p=3.6\times\:10^{-2}}K
p
=3.6×10
−2
Temperature (T) = 500K
R = 0.0831
To find:
Value of \sf{K_c}K
c
for the reaction at the same temperature.
Solution:
Given reaction:
\sf{N_2(g)+3H_2(g)\leftrightharpoons\:2NH_3(g)}N
2
(g)+3H
2
(g)⇋2NH
3
(g)
At first find the value of ∆n.
We know that,
∆n = Total moles of gas on the product side - Total moles of gas on the reactants side.
\implies⟹ ∆n = 2 - (1+3)
\implies⟹ ∆n = 2- 4
\implies⟹ ∆n = -2
We know that,
{\boxed{\sf{K_p=K_c(RT)^{\triangle\:n}}}}
K
p
=K
c
(RT)
△n
[ Put values ]
\begin{gathered} \implies \sf \: 3.6 \times {10}^{ - 2} = K_{c}(0.0831 \times 500)^{ \triangle \: - 2} \\ \\ \implies \sf \: K_{c} = \dfrac{3.6 \times {10}^{ - 2} }{0.0831 \times 500} \\ \\ \implies \sf \: K_{c} = 8.66 \times {10}^{ - 4} \end{gathered}
⟹3.6×10
−2
=K
c
(0.0831×500)
△−2
⟹K
c
=
0.0831×500
3.6×10
−2
⟹K
c
=8.66×10
−4
Therefore, the value of \sf{K_c}K
c
is \sf{8.66\times\:10^{-4}}8.66×10
−4
.