Answer:
178.55
Explanation:
176
×
0.05+
177
×
0.19
+
178
×
0.27
+
179
×
0.14
+
180
×
0.35
=
178.55
Explanation:
The given chemical reaction is:
![2Br^- (aq) + I_2(s) Br_2(l) + 2I^- (aq)](https://tex.z-dn.net/?f=2Br%5E-%20%28aq%29%20%2B%20I_2%28s%29%20%3C-%3E%20Br_2%28l%29%20%2B%202I%5E-%20%28aq%29)
![E^ocell=oxidation potential of anode + reduction potential of cathode\\](https://tex.z-dn.net/?f=E%5Eocell%3Doxidation%20potential%20of%20anode%20%2B%20reduction%20potential%20of%20cathode%5C%5C)
The relation between Eo cell and Keq is shown below:
![deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell](https://tex.z-dn.net/?f=deltaG%3D-RTlnK_e_q%5C%5Cdelta%3D-nFE%5Eo%20cell%5C%5C%3D%3EnFE%5Eo%20cell%3DRTlnK_e_q%5C%5ClnK_e_q%3D%5Cfrac%7BnF%7D%7BRT%7D%20E%5Eo%20cell)
The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,
![E^ocell= (-1.07+0.53)V\\=-0.54V](https://tex.z-dn.net/?f=E%5Eocell%3D%20%28-1.07%2B0.53%29V%5C%5C%3D-0.54V)
F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:
![ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3](https://tex.z-dn.net/?f=ln%20K_e_q%3D%5Cfrac%7B2mol%2A%2096485%20C%2Fmol%7D%7B8.314%20J.K%5E-%5E1.mol%5E-%5E1x298K%7D%20%5C%5C%5C%5ClnK_e_q%3D77.8%5C%5CK_e_q%3De%5E7%5E7%5E.%5E8%5C%5C%3D%3EK_e_q%3D6.13x10%5E3%5E3)
Answer:
Keq=6.13x10^33
Given:
<span> 2.1 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)
Required:
volume of CL2
Solution:
Use the ideal gas law
PV = nRT
V = nRT/P
V = (2.1 moles Cl2) (0.08203 L - atm / mol - K) (273K) / (1 atm)
V = 47 L</span>