A) \[2.66\,L\,\,{{\min }^{-1}}\,\text{at}\,\,\text{STP}\]
B) \[1.34\times {{10}^{-2\,}}\,\text{mol}\,\,{{\min }^{-1}}\]
C) \[6.96\times {{10}^{-2}}\,\text{mol}\,\,{{\min }^{-1}}\]
D) \[6.93\times {{10}^{-4}}\,\text{mol}\,\,{{\min }^{-1}}\]
Correct Answer: D
Solution :
[d] \[{{H}_{2}}{{O}_{2}}(aq)\to {{H}_{2}}O(aq)+\frac{1}{2}{{O}_{2}}(g)\] For a first order reaction \[k=\frac{2.303}{t}\log \frac{a}{(a-x)}\] Given a\[=0.5,(a-x)=0.125,t=50\,\,\min \] \[\therefore \,\,\,\,\,\,k=\frac{2.303}{50}\log \frac{0.5}{0.125}\] \[=2.78\times {{10}^{-2}}{{\min }^{-1}}\] \[r=k[{{H}_{2}}{{O}_{2}}]=2.78\times {{10}^{-2}}\times 0.05\] \[=1.386\times {{10}^{-3}}mol\,\,{{\min }^{-1}}\] Now \[-\frac{d[{{H}_{2}}{{O}_{2}}]}{dt}=\frac{d[{{H}_{2}}O]}{dt}=\frac{2d[{{O}_{2}}]}{dt}\] \[\therefore \]\[\frac{2d[{{O}_{2}}]}{dt}=-\frac{d[{{H}_{2}}{{O}_{2}}]}{dt}\] \[\therefore \]\[\frac{d[{{O}_{2}}]}{dt}=\frac{1}{2}\times \frac{d[{{H}_{2}}{{O}_{2}}]}{dt}\] \[\therefore \]\[\frac{d[{{O}_{2}}]}{dt}=\frac{1}{2}\times \frac{d[{{H}_{2}}{{O}_{2}}]}{dt}\] \[=\frac{1.386\times {{10}^{-3}}}{2}=6.93\times {{10}^{-4}}\,\text{mol}\,\,{{\min }^{-1}}\]
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