Consider the following reaction: H_2 (g) + I_2 (g) Equilibrium 2 HI (g) Complete the following...

Question

Question

science chemistry

Answer 1

Answer #1

Ans. Balanced reaction: H_2(g) + I_2(g) ------------> 2 HI_(g)

Equilibrium constant, Kc = [HI]² / ([H₂] [I₂]) - equation 1

Part A: Putting the values in equation 1-

Kc = (0.922)²/ (0.0355 x0.0388) = 0.850084 / 0.0013774 = 617.16

Hence, Kc = 617

Part B. Putting the values in equation 1-

90.6 = (0.385)²/ ([H₂] x 4.6 x10^-2)

Or, 90.6 x 0.046 [H₂] = 0.148225

Or, [H₂] = 0.148225 / 4.1676 = 0.04

Thus, [H₂] = 0.04 M

Part C: Putting the values in equation 1-

50.2 = ([HI])²/ (4.9 x 10^-2 x 4.64 x10^-2)

Or, ([HI])² = 50.2 x 0.0022736 = 0.11413472

Or, [HI] = (0.11413472)^1/2 or under root (0.11413472)

Hence, [HI] = 0.34

Thus, [HI] = 0.34

Answer 2

Similar Homework Help Questions

IV Consider the reaction: H_2 (g) + I_2 (g) = 2 HI(g) A reaction mixture in...

IV Consider the reaction: H_2 (g) + I_2 (g) = 2 HI(g) A reaction mixture in a 4.25 L flask at a certain temperature initially contains 0.668 g H_2 (g) and 86.9 g of I_2 (g). At equilibrium, the flask contains 78.4 g HI. Calculate K_c degree for this reaction at this temperature.
K_c for the reaction of hydrogen and iodine to produce hydrogen iodide, H_2 (g) + I_2...

K_c for the reaction of hydrogen and iodine to produce hydrogen iodide, H_2 (g) + I_2 (g) 2HI (g) is 54.3 at 430 degree C. Determine the initial and equilibrium concentration of HI if initial concentrations of H_2 and I_2 are both 0.10 M and their equilibrium concentrations are both 0.043 M at 430 degree C. [HI]_i = M [HI]_e = M
The equilibrium constant, K_c, for the following reaction is 55.6 at 698 K: H_2(g) + I_2(g)...

The equilibrium constant, K_c, for the following reaction is 55.6 at 698 K: H_2(g) + I_2(g) 2HI(g) Calculate the equilibrium concentrations of reactants and product when 0.351 moles of H_2 and 0.351 moles of I_2 are introduced into a 1.00 L vessel at 698 K. [H_2] = M [I_2] = M [HI] = M

K_c for the reaction of hydrogen and iodine to produce hydrogen iodide, H_2(g)+I_2(g) doubleheadarrow 2HI(g) is...

K_c for the reaction of hydrogen and iodine to produce hydrogen iodide, H_2(g)+I_2(g) doubleheadarrow 2HI(g) is 54.3 at 430 degree C. Determine the initial and equilibrium concentration of H_2and I_2 are both 0.11 M and their equilibrium concentrations are both 0.048 M at 430 degree
Part A, B, and C a Rev Consider the following reaction: H ( + 1, ()...

Part A, B, and C a Rev Consider the following reaction: H ( + 1, () 2 HI(g) Complete the following table. Assume that all concentrations are equilibrium concentrations in M. - Part A Find at 25°C Express your answer using three significant figures. T (C) H 1, HIK 25 0.0355 0.0388 0.922 - 340 - 4.60x10 PM 0.395 M 9.6 445 90x10-2M 4.72x10-2M - 502 V AE RO? You may want to reference (Pages 656 664) Section 15.8 while...
Consider the following reaction: H2 (g) + I2 (g) ⇌ 2 HI (g) Complete the following...

Consider the following reaction: H2 (g) + I2 (g) ⇌ 2 HI (g) Complete the following table. Assume that all concentrations are equilibrium concentrations in M. T(∘C) [H2] [I2] [HI] [Kc] 25 0.0355 0.0388 0.922 − 340 − 4.55×10−2 M 0.384 M 90.6 445 4.90×10−2 M 4.74×10−2 M − 50.2 Find Kc at 25 ∘C. Find [H2] at 340 ∘C. Find [HI] at 445 ∘C.

During an experiment, 0.986 mol H_2 and 0.493 mol I_2 were placed in a 1.95 L...

During an experiment, 0.986 mol H_2 and 0.493 mol I_2 were placed in a 1.95 L vessel where the following reaction came to equilibrium. H_2 (g) + I_2 (g) 2HI(g) For this reaction K_c = 49.5 at the temperature of the experiment. a) What were the equilibrium concentrations of H_2, I_2, and HI? [H_2] = [I_2] = [HI] =
At a particular temperature, K = 1.00 Times 10^2 for the following reaction. H_2(g) + I_2(g)...

At a particular temperature, K = 1.00 Times 10^2 for the following reaction. H_2(g) + I_2(g) 2 HI(g) IN an experiment, 1.39 mol H_2, 1.39 mol HI are introduced intoa 1.00-L, container, Calculate the concentrations of all species when equilibrium is reached. H_2 M I_2 M HI M
At 400 K, an equilibrium mixture of H_2, I_2, and HI consists of 0.068 mol H_2,...

At 400 K, an equilibrium mixture of H_2, I_2, and HI consists of 0.068 mol H_2, 0.075 mol I_2, and 0.13 mol HI in a1.00-L flask. What is the value of K_p, for the flowing equilibrium? (R = 0.0821 L middot atm (K - mol)) 2HI_(g) Rightwardsharpoonoverleftwardsharpoon H_2 (g) + I_2 (g) A) 0.039 B) 3.4 C) 26 D) 0.29 E) 8.2 If K = 0.150 for A_2 + 2B Rightwardsharpoonoverleftwardsharpoon 2AB, what is the value of K for the...

The equation for the formation of hydrogen iodide from H_2 and l_2 is: H_2(g) + I_2(g)...

The equation for the formation of hydrogen iodide from H_2 and l_2 is: H_2(g) + I_2(g) 2 HI(g) The value of K_p for the reaction is 69.0 at 790.0 degree C. What is the equilibrium partial pressure of HI in sealed reaction vessel at 790.0 degree C if the initial partial pressures of H_2 and l_2 are both 0.1800 atm an initially there is no HI present?