Question

\(NO_2\) required for a reaction is produced by decomposition of \(N_2 O_5\) in \(CCl_4\) as by equation

\(2N_2 O _ {5 (g)} \rightarrow 4 NO _ {2 (g)} +O_{2(g)}\)

The initial concentration of \(N_2 O _5\) is 3 mol \(L^{-1}\) and it is 2.75 mol \(L^{-1}\) after 30 minutes.

The rate of formation of \(NO_2\) is x \(\times 10^{-3}\) mol \(L^{-1} \,min ^{-1},\) value of x is ________.

Answer 1

Correct answer is : 17

Rate of reaction (ROR)

Rate of formation of \(NO _2 = \frac {\triangle [NO_2 ]}{\triangle t} = 4 \times ROR \)

\(= \frac {4}{240} = 16.66 \times 10 ^{-3}\) \(mol\,L^{-1}\) \(min ^{-1} \simeq 17 \times 10 ^{-3}.\)