 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/4/c/i/solution

Convert the supplied concentrations to

[NO] = 1.00 \times 1 0^{- 7} M

and

[O_{2}] = 3.00 \times 1 0^{- 5} M

. Then

[ONOO]_{ss} = \frac{189 ( 1.00 \times 1 0 ^{- 7} ) ( 3.00 \times 1 0 ^{- 5} )}{6500 + ( 1.0 \times 1 0 ^{9} ) ( 1.00 \times 1 0 ^{- 7} )} = 8.59 \times 1 0^{- 14} M .

The combined loss coefficient is

6600 s^{- 1}

. Neglecting the smaller NO-dependent loss would give

8.72 \times 1 0^{- 14} M

, a close low-NO approximation, but the full expression is just as easy to evaluate.

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