please help with highlighted portion v1.56-2.00-r A positive number (1.56) has a

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please help with highlighted portion

v1.56-2.00-r A positive number (1.56) has a positive and a negative root, b ny chemical meaning, so we ignore the neg this case, only the ot: positive root has a or 2.50 1.25 therefore x=1.11M 2.50= 2.25x So Calculating equilibrium concentrations: [CO] = [H2O] = 2.00 M-= 2.00 M- [CO] = [H,] = x 1.11 M 1.11 M 0.89 M Check Given the intermediate size of K, (1.56), it makes sense that the chang the t-only reactants were initially present, so x has a negative roducts. Also check that the equilibriu in concentration are moderate. It's a good idea to check that the sign of x in reaction table is corre sien for reactants and a positive sign for p concentrations give the known ons give the knownK (0.89%0.89) FOLLOW-UP PROBLEM 17.8 The decomposition of HI at low temperature was studied by injecting 2.50 mol of HI into a 10.32-L vessel at 25°C. What is [H2] at equilibrium for the reaction 2HI(g) =-H2(g) + 12(g); Kc = 1 .26× 10-39 Using the Quadratic Formula to Find the Unknown The shortcut we used to simplify the math in Sample Problem 17.8 is a special case that occurred because we started with equal concentrations of the reactants, but tynicallu ent concentrations of reactanto

Explanation / Answer

concentration of HI = 2.50 / 10.32 = 0.242 M

2 HI (g)   <------------> H2 (g) + I2 (g)

0.242                           0              0

0.242 - 2x                     x               x

Kc = x^2 / (0.242 - 2x)^2

1.26 x 10^-3 = x^2 / (0.242 - 2x)^2

x / 0.242 - 2x = 0.0355

x = 8.59 x 10^-3 - 0.071x

x = 0.00802 M

concentration of H2 = 0.00802 M

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