Hydrogen iodide decomposes according to the reaction?
..2 HI (g) -->.. H2 (g).. + ..I2 (g) 0.0244 ...... 0.00623 .... 0.00414 0.00312 .... 0.00156 .... 0.00156 ---------------------------------------... 0.02128 .... 0.00467 .... 0.00258 so, concentrations of HI (g) = 0.02128 mol / 1.5 L = 0.014 M concentrations of I2 (g) = 0.00258 mol / 1.5 L = 0.00172 M
4) Hydrogen iodide decomposes according to the equation 2HI(g) --><-- H2(g) + I2(g), for which K= 0.0156 at 40?
2HI(g) <===> H2(g) + I2(g) ICE table to tract concentrations @ equilibrium: . . . . . . HI. . . . . . . H2. . . . . I2 I. . . . 0.275. . . . . . 0. . . . . . 0 C. . . . -x. . . . . . . . .+x. . . . . +x E. . . 0.275-2x. . . . .x. . . . . . x Kc = [H2]*[I2] / [HI]² Kc = x*x / (0.275 - 2x)² Kc = x² / (0.275 - 2x)² .....Take the square root of both sides 0.0156 = x / 0.275 - 2x x = 0.027 From our ICE table, the concentration of H2 @ equilibrium was x. Thus, at equilibrium, the concentration of H2 = 0.027 M, none of these answers match exactly, but C is closest.
Hydrogen iodide decomposes according to the following reaction.2 HI(g) H2(g) + I2(g)?
A sealed 1.5 L container initially holds 0.00608 mol H2, 0.00434 mol I2, and 0.0217 mol HI at 703 K. When equilibrium is reached, the equilibrium concentration of H2(g) is 0.00423 M. What are the equilibrium concentrations of HI(g) and I2(g)?
Hydrogen iodide decomposes according to: 2HI(g) <----> H2(g)+ I2(g), Kc=0.0156 at 400degree Cel?
If 0.66 mol of HI injected into 2 L, the initial conc is 0.33 mol/L. Assume x mol/L of H2 is formed. From the equation, x mol/L of I2 is formed. The remaining conc of HI would be 0.33 -2x mol/L Substituting into the equilibrium constant yields x . x /(0.33 - 2x)^2 =0.0156 Taking square roots of both side and solving for x gives x=0.03297 mol/L Therefore HI 0.2641 mol/l H2 0.03297 mol/L I2 0.03297 mol/L You will need to round to the appropriate no of sig figures. Your data suggests 3
What happens when hydrogen peroxide reacts with potassium iodide?
When hydrogen peroxide is added to potassium iodide in neutral solution, the potassium iodide acts as a catalyst in the decomposition of hydrogen peroxide into water and oxygen.The reaction occurs in two stages, via an intermediate oxyiodide (OI-) ion;-H2O2 (aq) + I- (aq) --> H2O (l) + OI- (aq) H2O2 (aq) + OI- (aq) --> H2O (l) + O2 (g) + I- (aq) Since the iodide ion is not consumed by the reaction, it is classed as a catalyst.In acid solution there is a different reaction. In the presence of H+ ions, the potassium iodide is oxidised by the hydrogen peroxide into elemental iodine, and the hydrogen peroxide is reduced to water;-2KI (aq) + 2H+ (aq) + H2O2 (aq) <---> I2 (s) + 2H2O (l) + 2K+ (aq)This reaction is reversible and therefore, unless the iodine is removed from the system, an equilibrium will be reached which will depend on temperature and the initial concentrations of the reactants.
How do you write a balanced equation for the decomposition of sodium chlorate?
NaClO3—————————→ NaCl + 3/2O2
Hydrogen iodide can be synthesized according to the reaction?
[H2] = 0.0112 mol / 2 = 0.0056 M [I2] = 0.0018/ 2 = 0.00090 M In this case Kp = Kc in fact since Kp = Kc (RT)^delta n delta n = 2 - 1 - 1 = 0 54.5 = [HI]^2 / 0.0056 x 0.00090 [HI] = 0.0166 M [H2] = 0.364 / 2 = 0.182 M Q = (0.0166)^2 / 0.182 x 0.00090 = 1.68 Q < Kc so the reaction will shift on the right
What is the reaction between Calcium Carbonate and Hydrochloric Acid?
When hydrochloric acid reacts with any carbonates/hydrogen carbonates the products formed are metal chloride , water and carbon dioxide.Since HCl decomposes salts of weaker acids.So the equation of the reaction between calcium carbonate and HCl is:CaCO3+2HCl =CaCl2+2H2O+CO2This reaction is also known as double decomposition reaction.