| The initial concentration of \[{{N}_{2}}{{O}_{5}}\]is 3.00 \[\text{mol}\,{{\text{L}}^{-1}}\] and it is 2.75 \[\text{mol}\,{{\text{L}}^{-1}}\] after 30 minutes. The rate of formation of \[N{{O}_{2}}\]is : |
A) \[2.083\times {{10}^{3}}mol\text{ }{{L}^{1}}mi{{n}^{1}}\]
B) \[4.167\times {{10}^{3}}mol\text{ }{{L}^{1}}mi{{n}^{1}}\]
C) \[8.333\times {{10}^{3}}mol\text{ }{{L}^{1}}mi{{n}^{1}}\]
D) \[1.667\times {{10}^{2}}mol\text{ }{{L}^{1}}mi{{n}^{1}}\]
Correct Answer: D
Solution :
\[2{{N}_{2}}{{O}_{5}}(g)\xrightarrow[{}]{{}}4N{{O}_{2}}(g)+{{O}_{2}}(g)\] \[t=0\,\,\,\,\,\,\,\,\,\,3.0M\] \[t=30\,\,\,\,\,\,\,\,\,\,2.75M\] \[\frac{-\Delta [{{N}_{2}}{{O}_{5}}]}{\Delta t}=\frac{0.25}{30}\] \[\frac{1}{2}\times \frac{-\Delta [{{N}_{2}}{{O}_{5}}]}{\Delta t}=\frac{1}{4}\times \frac{\Delta [N{{O}_{2}}]}{\Delta t}\] \[\frac{\Delta [N{{O}_{2}}]}{\Delta t}=\frac{0.25}{30}\times 2=1.66\times {{10}^{-2}}M/\min \]
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