JEE MAIN - Chemistry (2019 - 12th April Evening Slot - No. 10)
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5(g) $$ \to $$ 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :
2N2O5(g) $$ \to $$ 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :
2.083 × 10–3
mol L–1
min–1
8.333 × 10–3
mol L–1
min–1
4.167 × 10–3
mol L–1
min–1
1.667 × 10–2
mol L–1
min–1
Explanation
Rate of disappearance of N2O5 = $$ - {{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}}$$ = $$ - {{\left[ {2.75 - 3} \right]} \over {30}}$$ = $$ + {1 \over {120}}$$ mol L-1 min-1
Rate of formation of NO2 = $$ + {{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$
As $$ - {1 \over 2}{{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = + {1 \over 4}{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$
$$ \Rightarrow $$ $${{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$ = $$ + {1 \over {120}} \times {4 \over 2}$$ = 0.01667 = 1.667 × 10–2 mol L–1 min–1
Rate of formation of NO2 = $$ + {{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$
As $$ - {1 \over 2}{{\Delta \left[ {{N_2}{O_5}} \right]} \over {\Delta \left[ t \right]}} = + {1 \over 4}{{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$
$$ \Rightarrow $$ $${{\Delta \left[ {N{O_2}} \right]} \over {\Delta \left[ t \right]}}$$ = $$ + {1 \over {120}} \times {4 \over 2}$$ = 0.01667 = 1.667 × 10–2 mol L–1 min–1
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