A mixture of 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid \( 1.00-\mathrm{L} \) contalner and the system was allowed to come to equillbrium. The equilibrlum concentration of \( \mathrm{COBr}_{2} \) was \( 0.233 \mathrm{M} \). What is the value of \( K_{\mathrm{C}} \) for this reaction? \[ \mathrm{CO}(g)+\mathrm{Br}_{2}(g) \rightleftarrows \mathrm{COBr}_{2}(g) \] Multiple Choice 1.22 0.191 0.858 1.165 5.23
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The given reaction is: \[ \mathrm{CO}(g) + \mathrm{Br}_{2}(g) \rightleftarrows \mathrm{COBr}_{2}(g) \] Show more…
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A mixture 0.500 mole of carbon monoxide and 0.400 mole of bromine was placed into a rigid 1.00-L container and the system was allowed to come to equilibrium. The equilibrium concentration of COBr2 was 0.233 M. What is the value of Kc for this reaction? (6 points) (You need to set up an ice table) CO(g) + Br2(g) COBr2(g)
Ishu S.
Carbonyl bromide decomposes to carbon monoxide and bromine. $$ \operatorname{COBr}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ $\mathrm{K}$ is 0.190 at $73^{\circ} \mathrm{C} .$ Suppose you placed $0.500 \mathrm{mol}$ of $\mathrm{COBr}_{2}$ in a $2.00-\mathrm{I}$. flask and heated it to $73^{\circ} \mathrm{C}$ (Study Question 17 ). After equilibrium had been achieved, you added an additional 2.00 mol of $\mathrm{CO}$ (a) How is the equilibrium mixture affected by adding more CO? (b) When equilibrium is reestablished, what are the new equilibrium concentrations of $\mathrm{COBr}_{2}, \mathrm{CO},$ and $\mathrm{Br}_{2} ?$ (c) How has the addition of CO affected the percentage of $\mathrm{COBr}_{2}$ that decomposed?
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