A) \[0.3\,\log 2\]
B) \[3\ln \frac{1.6}{1.6-1.4}\]
C) \[\frac{\ln 2}{200}\]
D) \[\frac{3\ln 2}{10}\]
Correct Answer: D
Solution :
Let pressure due to to \[{{A}_{2}}B(g)\]be ?x? atm \[\therefore \]pressure due to inert gas = (2 - x) atm\[{{A}_{2}}B(g)\]\[\to \] | \[2A(g)\]\[+\] | \[B(g)\] | |
\[t=0\] | \[x\] | \[0\] | \[0\] |
\[t={{t}_{end}}\] | \[0\] | \[2x\] | \[x\] |
\[{{A}_{2}}B(g)\]\[\to \] | \[2A(g)\]\[+\] | \[B(g)\] | |
\[t=10\,\min \] | \[1.6-y\] | \[2y\] | y |
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