Indicate how the concentration of each species in the chemical equation will change to reestablish equilibrium after a decrease in pressure or volume. If there is no change in the concentration of the species, leave the box blank. CH, (g) + 2 H, S(g)=CS,(g) + 4 H, (g) after decreasing the pressure after decreasing the volume of the reaction vessel Answer Bank

0.400 + x Predict the equilibrium concentration of IBr in the reaction described below (for which Kc = 280 at the reaction temperature) by constructing an ICE table, writing an equilibrium expression for Kc, and solving for the equilibrium concentration. Complete Parts 1-3 before submitting your answer. ₂(g) + Br₂(g) 2 lBr(g) NEXT > In a 3.0 L container at high temperature, 0.400 mol of IBr is allowed to reach equilibrium. Fill in the ICE table with the appropriate value for each involved species to determine the partial pressures of all reactants and products. Where applicable, use the x variables to represent any unknown change in concentration. Initial (M) Change (M) Equilibrium (M) 0 0.400 + 2x 1 0.400 0.400- x 1₂(g) 0.133 0.400 - 2x 2 + Br₂(g) = +x 3 0.133 + x +2x 0.133 + 2x 2 IBr(g) -X 0.133 - x ORESET -2x 0.133 - 2x

A 13.4 L reaction vessel at 483°C contained 0.892 mol , H2, 0.620 mol I2, and 5.41 mol . HI. Assuming that the substances are at equilibrium, find the value of Kc at 483°C for the reaction of hydrogen and iodine to give hydrogen iodide. The equation is H2(g)+I2(g)->2HI(g) Kc=_________

True/False. Explain your answer in one sentence or less.   In an equilibrium process, the concentrations of products and of reactants are equal. The value of the equilibrium constant for a given reaction depends on the initial concentrations of reactants. Large value for equilibrium constant K means the reaction will be less complete at the equilibrium point. The concentration of a pure solid is left out of an equilibrium constant expression but a pure liquid is included.

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Consider the chemical reaction shown below with an equilibrium constant (K) of 3.0 at some temperature. You add 3.5 MA and 3.5 MB to a reaction vessel heated to this temperature. What will the reaction do to reach equilibrium? 2 A (g) 2 3 B (g) The reaction will form more reactants to reach equilibrium. O The reaction will form more products to reach equilibrium. O The answer depends on the volume of the reaction vessel. O Cannot be determined because the equilibrium constant changes over time. O The reaction is already at equilibrium.

Consider the equilibrium system described by the chemical reaction below. Determine the concentration of O, at equilibrium by writing the equilibrium constant expression and solving it. Complete Parts 1-2 before submitting your answer. = 2 H₂O(g) 2 H2(g) + O̟₂(g) 1 2 NEXT At this temperature, the Kc = 2.4 × 103 and the equilibrium concentrations of H2O and H2 are 0.11 M and 0.019 M, respectively. If [x] represents the equilibrium concentration of O2, set up the equilibrium expression for Kc to solve for the concentration. Each reaction participant must be represented by one tile. Do not combine terms. Кс = 2' = 2.4 × 103 > RESET [0.11] [0.019] 2[0.11] 2[0.019] [0.11]² [0.019]² [x] [x]² [2x] [2x]²

Enter your answer in the provided box. The equilibrium constant K. for the equation 2H2(g) + CO(g)=CH3OH(g) is 13 at a certain temperature. If there are 2.21 × 102 moles of H2 and 2.33 x 103 moles of CH;OH at equilibrium in a 4.31-L flask, what is the concentration of CO? M

Enter your answer in the provided box. Given the reaction 2HI(g) 2 H2(g) + I,(g) Kea = 1.0 × 101 eq A sample mixture of HI, H,, and I, at equilibrium, was found to have [H2] = 3.0 × 10¬' M and HI] = 7.6 x 10 M. Calculate the molar concentration of I, in the equilibrium mixture. [L] =

Consider the following reaction: N2 (g)  +  O2 (g) → 2 NO (g), Kc = 0.25 You start the reaction with 14.72 moles of NO (nitrogen monoxide) in a 2.0 L vessel. Calculate the concentration of O2 (in molarity) at equilibrium. (You do not need to solve a quadratic equation to solve this problem.)