At 1 atm and 25 °C, NO2 with an initial concentration of 1.00 M is 0.0033% decomposed into NO and O2. Calculate the value of the equilibrium constant for the reaction. 2NO2(g)⇌2NO(g)+O2(g)
Added by Eric T.
Step 1
00 M Percentage decomposed = 0.0033% Concentration decomposed = 0.0033% of 1.00 M = 0.0033 * 1.00 = 0.0033 M Concentration of NO2 after decomposition = 1.00 M - 0.0033 M = 0.9967 M ** Show more…
Show all steps
Your feedback will help us improve your experience
Ronald Prasad and 61 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
For the reaction 2NO(g) + O2(g) → 2NO2(g) at a certain temperature, the equilibrium concentrations were found to be [NO2] = 5.7 × 10–3 M, [O2] = 1.0 × 10–2 M, and [NO] = 2.0 × 10–3 M. Calculate the value of the equilibrium constant from these data (delete units).
Preeti K.
Calculate the value of the equilibrium constant for the reaction: N2(g) + 2O2(g) = 2NO2(g) if the concentrations of the species at equilibrium are [N2] = 0.0016, [O2] = 0.0027, and [NO2] = 0.0080.
Madhur L.
At $25^{\circ} \mathrm{C}, 0.0560 \mathrm{~mol} \mathrm{O}_{2}$ and $0.020 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}$ were placed in a $1.00 \mathrm{~L}$ container where the following equilibrium was then established. $$ 2 \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g) $$ At equilibrium, the $\mathrm{NO}_{2}$ concentration was $0.020 \mathrm{M}$. What is the value of $K_{\mathrm{c}}$ for this reaction?
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD