Solution Found!
An equilibrium mixture of the following reaction has [I2] = 0.0205 M at 1200 °C. What is
Chapter 15, Problem 58P(choose chapter or problem)
An equilibrium mixture of the following reaction has \(\left[\mathrm{I}_{2}\right]=0.0205 \mathrm{M}\) at \(1200^{\circ} \mathrm{C}\). What is the concentration of I?
\(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\)
\(K_{\mathrm{eq}}=1.1 \times 10^{-2} \text { at } 1200^{\circ} \mathrm{C}\)
Equation Transcription:
Text Transcription:
\left[\mathrm{I}_{2}\right]=0.0205 \mathrm{M}
1200^{\circ} \mathrm{C}
\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)
K_{\mathrm{eq}}=1.1 \times 10^{-2} \text { at } 1200^{\circ} \mathrm{C}
Questions & Answers
QUESTION:
An equilibrium mixture of the following reaction has \(\left[\mathrm{I}_{2}\right]=0.0205 \mathrm{M}\) at \(1200^{\circ} \mathrm{C}\). What is the concentration of I?
\(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\)
\(K_{\mathrm{eq}}=1.1 \times 10^{-2} \text { at } 1200^{\circ} \mathrm{C}\)
Equation Transcription:
Text Transcription:
\left[\mathrm{I}_{2}\right]=0.0205 \mathrm{M}
1200^{\circ} \mathrm{C}
\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)
K_{\mathrm{eq}}=1.1 \times 10^{-2} \text { at } 1200^{\circ} \mathrm{C}
ANSWER:Answer:
Step 1
Equilibrium equation :
I2(g)2I(g)
Expression for the equilibrium constant of the reaction :
Keq=[I]2/[I2].
[I2]= equilibrium concentrations of I2.
[I] =equilibrium concentration of I
Keq= equilibrium constant of the reaction .