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In the equation $2HI(l) -> H_2(g) + I_2(s)$, what are the reactants?
Of course, this is an equilibrium reaction: $2HI(g) rightleftharpoons I_2(g) + H_2(g)$, in which there is equality of forward and reverse rates. I could equally write, $I_2(g) + H_2(g)...
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How do you balance this chemical equation: $HI(g) → H_2(g) + I_2(g)$?
How could I remove the $1/2$ coefficient?
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The reaction?
H2(g)+ I2(g)---> 2HI(g)
may occur by the following mechanism:
k1(--->)
I2<=>2I (fast, equilibrium)
k-1(<---)
I + I + H2 ---> 2HI (slow)
k2
The reaction rate is a direct function of the reactants concentrations, aA+ bB → cC $"Rate" = k[A]^x [B]^y$ Where $a, b, c$: are the stoichiometric coefficients $x, y$:...
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The following reaction is at equilibrium at a particular temperature:$H_2(g) + I_2(g) -> 2HI(g)$. The $[H_2]_(eq) = .012 M, [I_2]_(eq) = .15 M$, and $[HI]_(eq) = .30 M$.What is the magnitude of $K_c$ for the reaction?
Even without doing any calculation, you should be able to look at the values given for the equilibrium concentrations of the three chemical species that take part in the reaction...
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What is the equilibrium constant expression for the following reaction? $FeO(s) + H_2(g) rightleftharpoons Fe(s) + H_2O(g)$?
Because ferrous oxide and iron are solids, they cannot express a concentration, and thus they do not appear in the equilibrium expression. The equilibrium depends solely on the concentrations, or...
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What is $K_(eq)$ for the reaction $N_2 + 3H_2 rightleftharpoons 2NH_3$ if the equilibrium concentrations are $[NH_3] = 3 M$, $[N_2] = 2 M$, and $[H_2] = 1 M$?
Use the definition of $K_(eq)$. $K_(eq) = (["products"]^(d,e,f, . . . ))/(["reactants"]^(a,b,c,. . . ))$ $= (["NH"_3]_(eq)^2)/(["H"_2]_(eq)^3["N"_2]_(eq))$ Just make sure you remember to use stoichiometric coefficients correctly. $3H_2$...
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For the reaction below, $K_p=81.9$. A reaction mixture contains $P_(I_2)=0.114$ atm, $P_(Cl_2)=0.102$ atm, and $P_(ICl)=0.355$ atm. Find the equilibrium values for $P_(I_2)$, $P_(Cl_2)$, and $P_(ICl)$?
Well, I would never recommend this "trial-and-error" method... I don't see why you should ever "have" to do it... just use the quadratic formula on the resulting quadratic equation. I...
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In an experiment, 0.200 mol H2 and 0.100 mol I2 were placed in a 1.00 L vessel where the following equilibrium was established: H2 + I2 <-----> 2HI For this reaction, Kc = 49.5, What were the equilibrium concentration for H2, I2 and HI?
The equilibrium concentrations for all the species involved in the reaction are as follows: $[HI] = "1.87 M"$ $[H_2] = "0.107 M"$ $[I_2] = "0.00660 M"$ So, start with the...
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1. For the reaction, $PCl_(3(g))+Cl_(2(g)) rightleftharpoons PCl_(5(g))$, $K_c = 96.2 $ at 400 K. If the initial concentrations are 0.22 mol/L of $PCl_3$ and 0.42 mol/L of $Cl_2$, what are the equilibrium concentrations of all species?
So, you know that you're dealing with an equilibrium reaction that has its equal to 96.2 at a certain temperature. The fact that $K_c$ is larger than one tells...
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Which process represents oxidation with respect to the metal centre?
$"A. "Fe(s)+2HCl(aq) rarr FeCl_2(aq) + H_2(g)$
$"B. "FeCl_2(s)+H_2(aq) rarr Fe(s) + 2HCl(aq)$
$"C. "Fe(NO_3)_2(aq)+2HCl(aq) rarr FeCl_2(aq) + 2HNO_3(aq)$
$"D. "Fe_2O_3(s) + H_2(g)$
Iron undergoes the redox process: $Fe^0 rarrFe^(2+)$. The hydrogen ion is reduced to dihydrogen gas. Oxidation: $Fe rarr Fe^(2+) + 2e^-$ Reduction: $H^+ + e^(-) rarr 1/2H_2uarr$
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