<u>Answer:</u> The value of
for
reaction is ![5.13\times 10^2](https://tex.z-dn.net/?f=5.13%5Ctimes%2010%5E2)
<u>Explanation:</u>
We are given:
Initial moles of nitrogen gas = 1.30 moles
Initial moles of hydrogen gas = 1.65 moles
Equilibrium moles of ammonia = 0.100 moles
Volume of the container = 1.00 L
For the given chemical equation:
![N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)](https://tex.z-dn.net/?f=N_2%28g%29%2B3H_2%28g%29%5Crightleftharpoons%202NH_3%28g%29)
<u>Initial:</u> 1.30 1.65
<u>At eqllm:</u> 1.30-x 1.65-3x 2x
Evaluating the value of 'x'
![\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol](https://tex.z-dn.net/?f=%5CRightarrow%202x%3D0.100%5C%5C%5C%5C%5CRightarrow%20x%3D0.050mol)
The expression of
for above equation follows:
![K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
Equilibrium moles of nitrogen gas = ![(1.30-x)=(1.30-0.05)=1.25mol](https://tex.z-dn.net/?f=%281.30-x%29%3D%281.30-0.05%29%3D1.25mol)
Equilibrium moles of hydrogen gas = ![(1.65-x)=(1.65-0.05)=1.60mol](https://tex.z-dn.net/?f=%281.65-x%29%3D%281.65-0.05%29%3D1.60mol)
Putting values in above expression, we get:
![K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%280.100%29%5E2%7D%7B1.25%5Ctimes%20%281.60%29%5E3%7D%5C%5C%5C%5CK_c%3D1.95%5Ctimes%2010%5E%7B-3%7D)
Calculating the
for the given chemical equation:
![2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)](https://tex.z-dn.net/?f=2NH_3%28g%29%5Crightleftharpoons%20N_2%28g%29%2B3H_2%28g%29)
![K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2](https://tex.z-dn.net/?f=K_c%27%3D%5Cfrac%7B1%7D%7BK_c%7D%5C%5C%5C%5CK_c%27%3D%5Cfrac%7B1%7D%7B1.95%5Ctimes%2010%5E%7B-3%7D%7D%3D5.13%5Ctimes%2010%5E2)
Hence, the value of
for
reaction is ![5.13\times 10^2](https://tex.z-dn.net/?f=5.13%5Ctimes%2010%5E2)