Consider the reaction: CO(g)+H2 O(g) ⇌CO2(g)+H2(g) Kp=0.0611 at

Consider the reaction: CO(g)+H2 O(g) ⇌CO2(g)+H2(g) ...

Consider the reaction: $$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$$ $$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K_{\mathrm{p}}=0.0611 \mathrm{at} 2000 \mathrm{K} $$ A reaction mixture initially contains a CO partial pressure of 1344 torr and a $\mathrm{H}_{2} \mathrm{O}$ partial pressure of 1766 torr at 2000 $\mathrm{K}$ . Calculate the equilibrium partial pressures of each of the products.

Consider the reaction:
$$\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)$$
$$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad K_{\mathrm{p}}=0.0611 \mathrm{at} 2000 \mathrm{K}
$$
A reaction mixture initially contains a CO partial pressure of 1344 torr and a $\mathrm{H}_{2} \mathrm{O}$ partial pressure of 1766 torr at 2000 $\mathrm{K}$ . Calculate the equilibrium partial pressures of each of the products.

Key Concepts

ICE Table Method

The ICE table method (which stands for Initial, Change, Equilibrium) is a valuable tool for solving equilibrium problems. It involves organizing the initial conditions of the system, specifying the changes occurring as the system reaches equilibrium, and then calculating the equilibrium values. This structured approach helps to clearly relate stoichiometric relationships to the equilibrium constant expression.

Stoichiometry in Equilibrium Problems

Stoichiometry in equilibrium problems involves using the balanced chemical equation to relate the proportional changes in reactants and products as the system approaches equilibrium. The mole ratios provided by the equation guide the determination of how the concentration or partial pressure changes of one species correspond to those of another, ensuring that the conservation of mass is satisfied in the calculations.

Chemical Equilibrium

Chemical equilibrium refers to the state reached in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant concentrations or partial pressures of all species. It marks the point where the system shows no net change despite ongoing microscopic reactions.

Equilibrium Constant (Kp)

The equilibrium constant in terms of partial pressures, denoted Kp, is a numerical value that describes the ratio of the partial pressures of the gaseous products to those of the reactants at equilibrium, each raised to the power of their coefficients from the balanced chemical equation. It is temperature-dependent and indicates the extent to which a reaction favors products or reactants at equilibrium.

Partial Pressures

Partial pressure refers to the pressure exerted by an individual gas in a mixture, and it is particularly important in equilibrium calculations for gaseous reactions. It is used in place of concentration when applying the equilibrium constant expression for gases, as it reflects the effective 'concentration' of each gas in a pressure-based system.

Transcript

00:01 And we are told that the equilibrium constant of this equation is 0 .611.

00:06 We're also told that the initial concentrations of co and h2o are 1 ,34 tor and 17662 respectively.

00:15 So we are asked to determine the equilibrium concentrations of each of the reactants and products.

00:24 So the very first thing we're going to do before anything else, we need everything in atmospheres.

00:30 We cannot work in tor.

00:32 So we're going to divide these by 760 as one atmosphere is 760 tor.

00:41 So same here, we're going to divide by 760.

00:44 So we should wind up getting 1 .8 atmospheres of co, and we should get 2 .3 atmospheres of water.

00:55 So those are our initial values here.

00:57 So we have 1 .8 atmospheres and then we have 2 .3 atmospheres.

01:05 Initially we have no co2 and we have no hydrogen.

01:11 So we are going to just start right off the bat.

01:16 We are adding and then subtracting x.

01:23 This is going to be our overall change.

01:25 These all have coefficients of one.

01:28 So it's just going to be 1x for each one.

01:30 And then we're going to fill in for the equal of ram constant.

01:34 So for co we're going to have 1 .8 minus x, h2o will be 2 .3 minus x.

01:43 Co2 will just be x and h2 will just be x.

01:46 So next we're going to write out the equilibrium constant expression so that we can fill in with our equal of room constant values.

01:54 So k is equal to our products over our reactants.

01:59 So we have co2 here.

02:02 Hydrogen co and h2o.

02:11 So now we just need to fill in.

02:14 So our kp is 0 .0611 equals 1 .8 minus x times i grabbed the wrong value should be x times x divided by 1 .8 minus x times x times 2 .3 minus x...

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