Question

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]​

Answer 1

Answer:

Value of $\sf{K_c}$ = $\sf{8.66\times\:10^{-4}}$.

Explanation:

Given:

• Reaction : $\sf{N_2+3H_2\leftrightharpoons\:2NH_3}$
• $\sf{K_p=3.6\times\:10^{-2}}$
• Temperature (T) = 500K
• R = 0.0831

To find:

• Value of $\sf{K_c}$ for the reaction at the same temperature.

Solution:

Given reaction:

• $\sf{N_2(g)+3H_2(g)\leftrightharpoons\:2NH_3(g)}$

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,

${\boxed{\sf{K_p=K_c(RT)^{\triangle\:n}}}}$

• [ Put values ]

$:\implies \sf \: 3.6 \times {10}^{ - 2} = K_{c}(0.0831 \times 500)^{ \triangle \: - 2} \\ \\ :\implies \sf \: K_{c} = \dfrac{3.6 \times {10}^{ - 2} }{0.0831 \times 500} \\ \\ :\implies \sf \: K_{c} = 8.66 \times {10}^{ - 4}$

Therefore, the value of $\sf{K_c}$ is $\sf{8.66\times\:10^{-4}}$.