Answer:

Explanation:
Hello,
In this case, the law of mass action for the first reaction turns out:
![Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%3D1.27)
Now, for the second reaction is:
![Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D)
Therefore, by applying square root for the first reaction, one obtains:
![\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Csqrt%7BKc%7D%20%3D%5Csqrt%7B%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%7D%20%3D%5Csqrt%7B1.27%7D)
![\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5BAsH_3%5D%5E2%7D%20%7D%7B%5Csqrt%7B%5BAs%5D%5E2%7D%20%5Csqrt%7B%5BH_2%5D%20%5E3%7D%20%7D%20%3D%5Csqrt%7B1.27%7D)
![\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}](https://tex.z-dn.net/?f=%5Csqrt%7B1.27%7D%3D%5Cfrac%7B%5BAsH_3%5D%7D%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D)
Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:
![Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1.27%7D%20%7D%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D%3D0.887)
Best regards.
Zinc, selenium, antimony, arsenic, lead, and indium.
Answer:
Option A
Explanation:
Standard molar entropy determines the randomness of a mole of substance. Thus, the larger and complex a molecule is, the larger will be its molar entropy.
The molar entropy of gases is higher than liquids which is turn is higher than solid.
Here , NO2 is a gas with higher complex structure (trigonal planar) than N2 (linerar), N2O (linear) and N2O2 (linear)
Hence, NO2(g) will have greatest standard molar entropy
Option A is correct