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Magazine Chapter 13: Q81E (page 760)
Calculate the equilibrium concentrations of NO, O2, and NO2 in a mixture at 250 掳C that results from the reaction of 0.20 M NO and 0.10 M O2. (Hint: K is large; assume the reaction goes to completion then comes back to equilibrium.)
\(2NO(g) + {O_2}(g) \rightleftharpoons 2N{O_2}(g)\quad {K_c} = 2.3 \times 1{0^5}\;at\;25{0^o}C\)
Short Answer
The equilibrium concentrations of NO, O2,and NO2 are
\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,[{\rm{NO}}] = 7.04 \times {{10}^{ - 3}}M}\\{\,\,\,\,\,\,\,\,\,\,\left[ {{{\rm{O}}_2}} \right] = 3.52 \times {{10}^{ - 3}}M}\\{[{\rm{NO}}] = 0.193{\rm{M}}}\end{array}\).
Step by step solution
Determine change in concentration:
Given information:
\(2{\text{NO}}({\text{g}}) + {{\text{O}}_2}({\text{g}}) \rightleftharpoons 2{\text{N}}{{\text{O}}_2}({\text{g}})\)
- The concentration of NO is 0.20 M.
- The concentration of O2 is 0.10 M.
- The equilibrium constant is \({K_c} = 2.3 \times {10^5}\).
We have to find the equilibrium concentrations of NO, O2, and NO2.
- We will assume that the volume of a solution is 1 L, hence the number of moles of NO is 0.20 mol and the number of moles of O2 is 0.10 mol.
Since 2 moles of NO reacts with 1 mole of O2, to produce 2 moles of NO2, 0.20 mole of NO will exact with 0.10 mole of O2, and produce 0.20 moles of NO2(0.20 M).
We have to determine the value of x,
\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{K_c} = \frac{{{{\left[ {{\rm{N}}{{\rm{O}}_2}} \right]}^2}}}{{{{[NO]}^2} \times \left[ {{O_2}} \right]}}}\\{2.3 \times {{10}^5} = \frac{{{{(0.2 - 2x)}^2}}}{{{{(2x)}^2} \times x}}}\end{array}\)
Since \({K_c}\)is too large, we will neglect the change in concentration of \({\rm{N}}{{\rm{O}}_2}\), and get
\(\begin{array}{*{20}{c}}{2.3 \times {{10}^5} = \frac{{{{(0.2)}^2}}}{{{{(2x)}^2} \times x}}}\\{\,\,\,\,\,\,\,\,\,\,\,4{x^3} = \frac{{{{(0.2)}^2}}}{{2.3 \times {{10}^5}}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{x^3} = 4.35 \times {{10}^{ - 8}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3.52 \times {{10}^{ - 3}}{\rm{M}}}\end{array}\)
Determine equilibrium concentration of all species:
The change in the concentration obtained \(x = 3.52 \times {10^{ - 3}}{\rm{M}}\)
Therefore, the equilibrium concentrations of NO, O2, and NO2 are
\(\begin{array}{*{20}{c}}{\,\,[{\rm{NO}}] = 2{\rm{x}} = 7.04 \times {{10}^{ - 3}}{\rm{M}}}\\{\left[ {{{\rm{O}}_2}} \right] = {\rm{x}} = 3.52 \times {{10}^{ - 3}}{\rm{M}}}\\{\,\,\,\,\,\,\,[{\rm{NO}}] = 0.2{\rm{M}} - 2{\rm{x}} = 0.193{\rm{M}}}\end{array}\).
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