When 46g of I2 and 1g of H2 are heated to equilibrium at 450°C ,the equilibrium mixture contains 1.9g I2. How many moles of each gas present at equilibrium?Determine Kc and Kp for this reaction at the same temperature?

Question

Cameron Andrews · Accepted Answer

Here's what I got.

Start by writing the balanced chemical equation for this equilibrium reaction #"I"_text(2(g]) + "H"_text(2(g]) rightleftharpoons color(red)(2)"HI"_text(2(g])# Notice that you have a #1:1# mole ratio between the two reactants and a #1:color(red)(2)# mole ratio between the reactants and the product. Use the molar masses of hydrogen gas and iodine to determine how many moles of each you're adding to the container #46 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81color(red)(cancel(color(black)("g")))) = "0.1812 moles I"_2# and #1 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.0159color(red)(cancel(color(black)("g")))) = "0.4961 moles H"_2# You know that at equilibrium, the reaction vessel contains #"1.9 g"# of iodine. Calculate how many moles of iodine you have present at equilibrium #1.9 color(red)(cancel(color(black)("g"))) * "1 mole I"_2/(253.81 color(red)(cancel(color(black)("g")))) = "0.007486 moles I"_2# This tells you that a total of #n_"converted" = 0.1812 - 0.007486 = "0.1737 moles I"_2# have been converted to hydrogen iodide, #"HI"#. According to the #1:color(red)(2)# mole ratio that exists between iodine and hydrogen iodide, the reaction produced #0.1737 color(red)(cancel(color(black)("moles I"_2))) * (color(red)(2)" moles HI")/(1color(red)(cancel(color(black)("mole I"_2)))) = "0.3474 moles HI"# The #1:1# mole ratio that exists between hydrogen gas and iodine tells you that the reaction consumed equal numbers of moles of the two reactants, so at equilibrium you'll be left with #n_(H_2) = 0.4961 - 0.1737 = "0.3224 moles H"_2# Therefore, at equilibrium, the reaction vessel will contain - I'll leave the answers rounded to two sig figs, just for good measure #n_(I_2) = color(green)("0.0075 moles I"_2)# #n_(H_2) = color(green)("0.32 moles H"_2)# #n_(HI) = color(green)("0.35 moles HI")# By definition, the equilibrium constant for this reaction, #K_c#, which uses equilibrium concentrations, is equal to #K_c = (["HI"]^color(red)(2))/(["I"_2] * ["H"_2])# As you know, molarity is defined as moles of solute per liters of solution. In this case, the volume of the reaction vessel, let's say #V#, is the same for all three chemical species, so you can say that #["HI"] = "0.35 moles"/V# #["H"_2] = "0.32 moles"/V# #["I"_2] = "0.0075 moles"/V# This means that you have #K_c = ( 0.35)^2/color(red)(cancel(color(black)(V^2))) * color(red)(cancel(color(black)(V)))/0.32 * color(red)(cancel(color(black)(V)))/0.0075 = color(green)(51)# The relationship between #K_c# and #K_p# is given by the equation #color(blue)(K_p = K_c * (RT)^(Deltan))" "#, where #R# - the universal gas constant, equal to #0.0821("atm" * "L")/("mol" * "K")# #T# - the temperature at which the reaction takes place - expressed in Kelvin #Deltan# - the difference between the number of moles of gas found on the products' side and the number of moles of gas found on the reactants' side Notice that you reaction has a total of #color(red)(2)# moles of gas on the products' side, and #2# moles of gas, one from each reactant, on the reactants' side. This means that #Deltan = 0#, which implies that #K_p = K_c * (RT)^0# #K_p = K_c#

Cooper Adams · Answer

undefined To determine the moles of each gas present at equilibrium, we first need to calculate the moles of I2 and H2 initially and then use the given information to find the moles of each gas at equilibrium.

Given:
- Initial mass of I2 = 46 g
- Initial mass of H2 = 1 g
- Mass of I2 at equilibrium = 1.9 g

1. Calculate the moles of I2 initially:
   Moles of I2 = Mass of I2 / Molar mass of I2
               = 46 g / (2 * 126.9 g/mol)
               ≈ 0.182 moles

2. Calculate the moles of H2 initially:
   Moles of H2 = Mass of H2 / Molar mass of H2
               = 1 g / (2 * 1.008 g/mol)
               ≈ 0.496 moles

3. Calculate the change in moles of I2:
   Change in moles of I2 = Moles of I2(initial) - Moles of I2(at equilibrium)
                          = 0.182 moles - (1.9 g / (2 * 126.9 g/mol))
                          ≈ 0.182 moles - 0.0075 moles
                          ≈ 0.175 moles

4. Calculate the change in moles of H2:
   Change in moles of H2 = Moles of H2(initial) - (0 moles, as H2 is not mentioned in the equilibrium condition)
                          = 0.496 moles

5. Calculate the moles of each gas at equilibrium:
   Moles of I2 at equilibrium = Moles of I2(initial) - Change in moles of I2
                               ≈ 0.182 moles - 0.175 moles
                               ≈ 0.007 moles

Moles of H2 at equilibrium = Moles of H2(initial) - Change in moles of H2
                               ≈ 0.496 moles - 0 moles
                               ≈ 0.496 moles

Now, to determine Kc and Kp at the same temperature, we need the balanced equation for the reaction between I2 and H2. Assuming it forms HI:

$$I_2(g) + H_2(g) \rightleftharpoons 2HI(g)$$

Kc = [HI]^2 / ([I2] * [H2])
Kp = (p_HI)^2 / (p_I2 * p_H2)

We can use the concentrations of HI, I2, and H2 at equilibrium to calculate Kc and Kp. However, since the concentrations are not provided, we can't calculate the exact values without further information.