Verified Solution • 2025–26 Session moderate

QJEEM24Chc9fa: JEE Main 2024 Chemistry - Class 11 JEEM24Chc9fa Solution: For the given reaction, choose the correct expression of from…

For the given reaction, choose the correct expression of KC\mathrm{K}_{\mathrm{C}} from the following :-

Fe(aq)3++SCN(aq)−⇌(FeSCN)(aq)2+\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}

Multiple Choice Options

  • A KC=[Fe3+][SCN−][FeSCN2+]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}{\left[\mathrm{FeSCN}^{2+}\right]}
  • B KC=[FeSCN2+][Fe3+][SCN−]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
  • C KC=[FeSCN2+]2[Fe3+][SCN−]\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]^2}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
  • D KC=[FeSCN2+][Fe3+]2[SCN−]2\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^2\left[\mathrm{SCN}^{-}\right]^2}

Step-by-Step Verified Solution

amp;KC= Products ion conc.  Reactants ion conc. amp;KC=[FeSCN2+][Fe3+][SCN−]\begin{aligned}& \mathrm{K}_{\mathrm{C}}=\frac{\text { Products ion conc. }}{\text { Reactants ion conc. }} \\& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring