Start Learning Free

Before You Begin

Question #44847

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of benzenediazonium sulfate, \ce(C6H5N2)2SO4\ce{(C6H5N2)2SO4}, has an initial concentration of 0.05 M0.05 \text{ M}. In the presence of a specific catalyst at 300 K300 \text{ K}, it undergoes a zero-order decomposition:

\ce(C6H5N2)2SO4(aq)+2H2O(l)→2C6H5OH(aq)+2N2(g)+2H+(aq)+SO42−(aq)\ce{(C6H5N2)2SO4 (aq) + 2 H2O (l) \rightarrow 2 C6H5OH (aq) + 2 N2 (g) ^ + 2 H+ (aq) + SO4^{2-} (aq)}

Assume phenol (\ceC6H5OH\ce{C6H5OH}) remains completely unionized, \ceN2\ce{N2} gas escapes completely, and all strong electrolytes are completely dissociated. The osmotic pressure of the solution is observed to increase at a constant rate of 0.1476 atm min−10.1476 \text{ atm min}^{-1}. Calculate the zero-order rate constant kk for the decomposition of benzenediazonium sulfate.

(Given: R=0.082 L atm K−1 mol−1R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1})

Options

  • A.

    3.0×10−3 mol L−1 min−13.0 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

  • B.

    6.0×10−3 mol L−1 min−16.0 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

  • C.

    2.0×10−3 mol L−1 min−12.0 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

  • D.

    1.2×10−3 mol L−1 min−11.2 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Units of Rate Constant (k)|Zero Order Rate Integration|Molecular Mass Abnormalities

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of colligative properties and chemical kinetics.

The core trap involves how students handle the rate of change of a colligative property when the number of particles changes during a zero-order reaction.

Students typically memorize π=iCRT\pi = iCRT and might erroneously apply dπdt=iinitialRTk\frac{d\pi}{dt} = i_{initial}RTk (leading to Option C) or ignore the van't Hoff factor entirely dπdt=RTk\frac{d\pi}{dt} = RTk (leading to Option B).

The hidden variable is the 'differential particle flux' (Δn\Delta n).

Because the reaction is zero-order, the rate of change of osmotic pressure depends strictly on the net difference in the number of dissolved particles between products and reactants (Δn=+2\Delta n = +2), not the absolute initial or final van't Hoff factors.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptUnits of Rate Constant (k)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree