A) \[V{{m}^{-1}}\]
B) \[8\times {{10}^{7}}\]
C) \[5\times {{10}^{-11}}\]
D) \[8\times {{10}^{-11}}\]
Correct Answer: D
Solution :
| [d] Given, |
| \[\frac{dQ}{dt}\] |
| \[\frac{dQ}{dt}=\frac{KL({{T}_{1}}-{{T}_{2}})}{A}\] |
| \[\frac{dQ}{dt}=\frac{K({{T}_{1}}-{{T}_{2}})}{LA}\] (i) |
| Hydrolysis of \[N{{H}_{4}}Cl\] takes place as, |
| \[\frac{dQ}{dt}=KLA({{T}_{1}}-{{T}_{2}})\] |
| or\[\frac{dQ}{dt}=\frac{KA({{T}_{1}}-{{T}_{2}})}{L}\] |
| Hydrolysis constant, \[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R-{{i}_{1}}{{r}_{1}}=0\] (ii) |
| or\[{{\varepsilon }_{2}}-{{i}_{2}}{{r}_{2}}-{{\varepsilon }_{1}}-{{i}_{1}}{{r}_{1}}=0\] |
| From Eqs (i) and (iii) |
| \[-{{\varepsilon }_{2}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\]\[{{\varepsilon }_{1}}-({{i}_{1}}+{{i}_{2}})R+{{i}_{2}}{{r}_{2}}=0\] |
| \[4\times {{10}^{-8}}m.\] \[V{{m}^{-1}}\] |
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