When an equation is reversed in direction the new Keq is reciprocal of the old one.
When an equationis divided by a factor x , then the K' = K1/x
and if we multiply the equation by x, then K' = Kx
Thus
K for the given new equation ( which is reversed and divided by 2)
K' = 1/K1/2
= 1/(3.6x104)1/2
= 5.27x10-3
For the reaction H_2(g) + Br_2 (g) rightarrow 2HBr(g) K_p = 3.6 times 10^4 at 1496...
Consider the following reaction: 2HBr (g) rightarrow H_2 (g) + Br_2 (g) In the first 23.0 s of this reaction the of HBr dropped from 0.550 M to 0.457 M. Calculated the average rate of the reaction in this time interval. Express your answer using two significant figures. If the volume of the reaction in part (b) was 1.50 L. what amount of Br_2 (in moles) was formed during the 15.0 is of the reaction? Express your answer using two...
Consider the following reaction: H_2 (g) + Br_2 (g) rlhar 2HBr(g) At some temperature, K_p = 35.5. Within a container, P_H_2 = 5.00 atm, P_Br_2 = 0.700 atm, and P_HBr = 1.60 atm Determine Q_p for this reaction under these conditions. (Round Q_p to 3 sig figs). Q_p = In which direction will the reaction shift to reach equilibrium? (Round Q_p to 3 sig figs).
Be sure to answer all parts. The equilibrium constant K_c for the reaction H_2(g) + Br_2(g) = 2HBr(g) is 2.180 Times 10^6 at 730DegreeC. Starting with 5.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H_2, Br_2, and HBr at equilibrium. [H_2] = M [Br_2] = M [HBr] = M
The equilibrium constant K_c for the reaction H_2 (g) + Br_2(g) 2 HBr(g) is 2.180 times 10^6 at 730 degree C. Starting with 2.20 moles of HBr in a 21.6-L reaction vessel, calculate the concentrations of H_2, Br_2 and HBr at equilibrium. [H_2] = [Br_2] = [HBr] =
Consider the following reaction between carbon monoxide and hydrogen, CO(g) + 2 H_2(g) Equilibrium CH_3OH(l) K_p = 2.25 times 10^4 Give the K_p expression for this reaction Determine the equilibrium pressure of CO if the P(H_2) at equilibrium = 0.52 atm If the Q_p for a particular reaction mixture was measure as 2.30 times 10^-4, how will this reaction proceed?
For the reaction H2(g) + Br2(g) → 2HBr(g) Kp = 3.6 x 104 at 1494 K. What is the value of Kp for the following reaction at 1494 K? 42 H2(g) + / Br2(g) HBr(g) K". p Submit
For the reaction 2 HBr(g) + Cl_2(g) rightarrow 2 HCl(g) + Br_2(g) a. Write the equilibrium constant expression for the reaction. b. Using the following G degree values, calculate Delta G degree for the reaction. HBr(g) = -53.22 kJ/mol HCl(g) = -95.27 kJ/mol Cl_2(g) = 0 kJ/mol Br_2(g) = 3.14 kJ/mol c. Calculate the equilibrium constant K_eq be at 298 K. d. Does this equilibrium lie more with reactants or products?
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...
For the reaction, the value of K_p is 41.0 (at 400.0 K). and deltaH^n = -92.2 kJ N_2(g) + 3 H_2(g) rightarrow 2 NH_3(g) What is the value of K_p at 705.0 K? Collect and Organize Given the values of deltaH_n and K_p for the conversion of N_2 and H_2 to give NH_3 at one temperature, we will calculate the value of K_p at a higher temperature. Analyze What is the expression for the relationship between equilibrium constants at two...
K_p for NH_3 at 25 degree C N_2 (g) + 3 H_2(g) irreversible 2 NH_3 (g), Delta G degree = -31.0 kJ consider the galvanic cell that uses the reaction 2 Ag^+ (aq) plus Cu(s) rightarrow Cu^2+ (aq) + 2Ag (s) clearly sketch the experimental set-up, write down the anode and cathode half- give the shorthand notation for the cell For the following cell, write a balanced equation for the cell reaction and calc Delta G degree C: Pt(s) |H_2(1.0...