© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write the Balanced Decomposition Reaction
The given decomposition reaction is:
2\,\mathrm{N_2O_5(g)} \rightarrow 4\,\mathrm{NO_2(g)} + \mathrm{O_2(g)} .
From the problem statement:
Initial concentration of \mathrm{N_2O_5} : 3\,\mathrm{mol\,L^{-1}}
Concentration of \mathrm{N_2O_5} after 30 minutes: 2.75\,\mathrm{mol\,L^{-1}}
Time interval, \Delta t : 30\,\mathrm{min}
Step 2: Express the Rate in Terms of the Disappearance of \mathrm{N_2O_5}
For the reaction
2\,\mathrm{N_2O_5} \rightarrow 4\,\mathrm{NO_2} + \mathrm{O_2},
the rate of reaction ( \mathrm{ROR} ) can be written as:
\mathrm{ROR} = -\frac{1}{2}\,\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t},
where the factor -\tfrac{1}{2} accounts for the stoichiometric coefficient 2 in front of \mathrm{N_2O_5} and the negative sign indicates disappearance of reactant.
Step 3: Calculate the Change in Concentration of \mathrm{N_2O_5}
\Delta[\mathrm{N_2O_5}]
= [\mathrm{N_2O_5}]_{\text{final}} - [\mathrm{N_2O_5}]_{\text{initial}}
= 2.75 - 3.00
= -0.25\,\mathrm{mol\,L^{-1}}.
It is negative because the concentration of \mathrm{N_2O_5} decreases.
Step 4: Determine the Rate of Reaction (ROR)
Substituting into the rate expression:
\mathrm{ROR}
= -\frac{1}{2} \times
\frac{-0.25\,\mathrm{mol\,L^{-1}}}{30\,\mathrm{min}}
= \frac{0.25}{2 \times 30}\,\mathrm{mol\,L^{-1}\,min^{-1}}
= \frac{1}{240}\,\mathrm{mol\,L^{-1}\,min^{-1}}.
Numerically,
\frac{1}{240} \approx 0.00417\,\mathrm{mol\,L^{-1}\,min^{-1}}.
Step 5: Calculate the Rate of Formation of \mathrm{NO_2}
From the balanced equation, for every 2 moles of \mathrm{N_2O_5} decomposed, 4 moles of \mathrm{NO_2} are produced. Hence:
\mathrm{Rate~of~formation~of~NO_2}
= 4 \times \mathrm{ROR}.
Substituting
\mathrm{ROR} = \tfrac{1}{240}\,\mathrm{mol\,L^{-1}\,min^{-1}} :
\mathrm{Rate~of~formation~of~NO_2}
= 4 \times \frac{1}{240}\,\mathrm{mol\,L^{-1}\,min^{-1}}
= \frac{4}{240}\,\mathrm{mol\,L^{-1}\,min^{-1}}
= \frac{1}{60}\,\mathrm{mol\,L^{-1}\,min^{-1}}.
Numerically,
\frac{1}{60} = 0.01666\ldots\,\mathrm{mol\,L^{-1}\,min^{-1}}
= 16.66\times10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}}.
Step 6: Express the Value in the Form x \times 10^{-3}
In the form x \times 10^{-3}\,\mathrm{mol\,L^{-1}\,min^{-1}},
the rate of formation of \mathrm{NO_2} is approximately
16.66 \times 10^{-3} .
The nearest integer to 16.66 is 17, so the value of x is \boxed{17} .
