I know how to solve it but i want to know what if the equation HI (g) 1/2 H2(g) + 1/2 I2(g) did not have 1/2 instead it had 3 or 2 how would the calculation change?
The value of Keq for the equilibriumH2(g) + I 2(g)= 2 HI (g) is 794 at 25°C. At this temperature, what is the value of Keq for the equilibrium below?
HI (g) 1/2 H2(g) + 1/2 I2(g)
A.0.035 B.0.0013 C.28D.397E.1588
I know how to solve it but i want to know what if the equation HI...
The value of Keq for the equilibrium H2 (g) + I2 (g) ⇌ 2HI (g) is 794 at 25 °C. At this temperature, what is the value of Keq for the equilibrium below? HI (g) ⇌ 1/2 H2 (g) + 1/2 I2 (g) 1588 0.0013 397 28 0.035
Consider the equilibrium reaction. H2(g) + I2(g) equilibrium reaction arrow 2 HI(g) In this case, 1.000 M H2 reacts with 2.000 M of I2 at a temperature of 414°C. The value of Kc = 72. Determine the equilibrium concentrations of H2, I2, and HI. [H2] [I2] [HI]
How many moles of H2 and HI will be present at equilibrium if 0.8430 mol HI are placed into a 1-L flask and allowed to react at a temperature where for the following reaction takes place: 2HI(g) H2(g) + I2(g) K = 16.5 mol H2? mol HI?
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24. The reaction: H2 (g) + I2 (g) <--> 2 HI (g) has a Keq = 20.0 If the initial concentrations give a value less than 20, we can say: a. the reaction will move to the reagents to reach equilibrium b. by increasing the volume of the reaction container, the system will reach equilibrium c. reducing the volume causes the reaction to reach equilibrium d. the reaction will move towards the products to reach equilibrium
the equilibrium constant Kc for the equation 2hi= h2 + I2 at 425c is 1.84 what is the value of kc for the following equation 1/4 H2 +1/4 I2 = 1/2 HI
How many moles of HI will be present at equilbrium if 0.6160 mol HI are placed into a 1-L flask and allowed to react at a temperature where for the following reaction takes place: 2HI(g) H2(g) + I2(g) K = 28.3 0.286 mol H2 ___ ? mol HI
1/2 H2 (g) + 1/2 I2 (g) <-> HI (g) At a certain temperature, Kc of reaction is 400 If 0.025M of H2 and 0.025 M and 0.065 M of HI are in a 1.00 L container, what is the concentration of HI at equilibrium?
The reaction H2(g) + I2(g) → 2 HI (g) is first order in both hydrogen and iodine. It is therefore referred to as second order overall. Its rate constant for the formation of HI (g) at 400 ◦C is 2.34 × 10−2 · lit · mol−1 · sec−1 and its activation energy is 150 KJ/mol. Use the rate law to estimate how long it takes to form 0.1 mole of HI(g) if I start by putting 2 moles of H2...
. HI decomposes according to the equation 2 HI (g) → H2 (g) + I2 (g). The rate constant k for this reaction was measured at several different absolute temperatures T, and an Arrhenius plot of ln k vs. 1/T had a slope of -2.24 × 10^4 K. What is the activation energy Ea for this reaction? A. 1.86 × 105 J/mol B. 3.28 × 103 J/mol C. -1.86 × 105 J/mol D. -3.28 × 103 J/mol