1. Tardigrade
  2. Question
  3. Chemistry
  4. The equilibrium constant at 850 K for the reaction N2(g) + O2(g) <=> 2NO(g) is 0.5625. The equilibrium concentration of NO(g) is 3.0 × 10-3 M. If the equilibrium concentrations of N2(g) and O2(g) are equal, the concentration of N2(g) in M is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The equilibrium constant at $850\, K$ for the reaction $\ce{N_{2(g)} + O_{2(g)} <=> 2NO_{(g)}} $ is $0.5625$. The equilibrium concentration of $\ce{NO_{(g)} } $ is $3.0 \times 10^{-3} M$. If the equilibrium concentrations of $\ce{N_{2(g)}} $ and $\ce{O_{2(g)}}$ are equal, the concentration of $\ce{N_{2(g)}} $ in $M$ is

AP EAMCETAP EAMCET 2019

Solution:

Given,

$N _{2}(g)+ O _{2}(g) \rightleftharpoons 2 NO (g)$

i.e. equilibrium constant, $K_{C}=0.5625$

Equilibrium concentration of $[ NO ]=3 \times 10^{-3} M$

$\because K_{C}=\frac{[ NO ]^{2}}{\left[ N _{2}\right]\left[ O _{2}\right]}$

But, $\left[ N _{2}\right]=\left[ O _{2}\right]$ (Given)

$\therefore K_{C}=\frac{[ NO ]^{2}}{\left[ N _{2}\right]^{2}}$

or, $0.5625=\frac{\left[3 \times 10^{-3}\right]^{2}}{\left[ N _{2}\right]^{2}}$

or, $\left[ N _{2}\right]^{2}=\frac{\left[3 \times 10^{-3}\right]^{2}}{0.5625}$

$=\frac{9 \times 10^{-6}}{0.5625}$

Or, $\left[ N _{2}\right]^{2}=16 \times 10^{-6}$

or, $\left[ N _{2}\right]=4 \times 10^{-3}$