Test Index

CBSE Class 12 Chemistry 2018 Solved Paper

© examsnet.com
Question : 7 of 26
Marks: +1, -0
For the reaction
2N2O5(g)→4NO2(g)+O2(g)  ,   2 \mathrm{N}_2 \mathrm{O}_5(\mathrm{g}) \rightarrow 4 \mathrm{NO}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \; \text{, } \;
the rate of formation of NO2(g)\mathrm{NO}_2(\mathrm{g}) is 2.8×10−3 M s−12.8 \times 10^{-3} \ \mathrm{M} \ \mathrm{s}^{-1}.
Calculate the rate of disappearance of N2O5(g)\mathrm{N}_2 \mathrm{O}_5(\mathrm{g})
Solution:  
  Rate    =  14  ΔNO2Δ(t)=  12  Δ(N2O5)Δ(t)\; \text{Rate} \;\; = \; \frac{1}{4} \; \frac{\Delta N \mathrm{O}_2}{\Delta(t)} = \; \frac{1}{2} \; \frac{\Delta ( \mathrm{N}_2 \mathrm{O}_5 )}{\Delta(t)}
  14(2.8×10−3)  =−  12  Δ(N2O5)Δ(t)\; \frac{1}{4} (2.8 \times 10^{-3}) \; = - \; \frac{1}{2} \; \frac{\Delta ( \mathrm{N}_2 \mathrm{O}_5 )}{\Delta(t)}
Rate of disappearance of N2O5\mathrm{N}_2 \mathrm{O}_5 (−ΔN2O5Δ(t))1.4×10−3 M/s\left( - \frac{\Delta \mathrm{N}_2 \mathrm{O}_5}{\Delta(t)} \right) 1.4 \times 10^{-3} \ \mathrm{M} / \mathrm{s}
(Deduct half mark if unit is wrong or not written)
© examsnet.com
Go to Question:
Prev Question
Next Question