Asked by jude
Hydrogen iodide decomposes according to the reaction 2HI(g)H2(g) + I2(g) A sealed 1.50 L container initially holds .00623 moles of H2, .00414 moles of I2, and .0244 moles of HI at 703K. When equilibrium constant is reached, the concentration of H2(g) is .00467 M. What are the concentrations of HI(g) and I2(g)?
Answers
Answered by
DrBob222
....2HI ==> H2 + I2
I...mols/L..mols/L...mols/L. You are given mols and L, calculate M.
C.....-2x....x.....x
E.....M-2x..M+x...M+x
You are given M+x for H2, solve for x and evaluate the others. Post your work if you get stuck. Note my comments on your other posts. Multiple screen names are frowned on here.
I...mols/L..mols/L...mols/L. You are given mols and L, calculate M.
C.....-2x....x.....x
E.....M-2x..M+x...M+x
You are given M+x for H2, solve for x and evaluate the others. Post your work if you get stuck. Note my comments on your other posts. Multiple screen names are frowned on here.
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