Now, we have observed total pressure of 2. Hence the total pressure for the reaction is: The third row sums up the expressions of the first two rows to determine the partial pressures at equilibrium. Based on the stoichiometric coefficients, if the value of PO 2 increases by x, PH 2 will change by 2x. Now, we look at the balanced equation in order to describe how the partial pressures can change when the reaction reaches an equilibrium. It can also be possible that there are no units for Kc because the concentrations at the top and the bottom both cancel eachother out. Generally, the relation between K p and K c can be represented as: The units for Kc changes depending on the concentrations of each reactant and product in the equation.Kc is the equilibrium constant, in molarity, which depicts the ratio of the equilibrium concentrations of products over the concentrations of reactants.K p is the equilibrium constant in atmospheric pressure determined from the partial pressures of the equation of a reaction.Therefore, the K p of this reaction is 1.24 × 10 −1 We know that the relation between K p and K c is K p = K c (RT) \(\bigtriangleup \)n Therefore, we can proceed to find the Kp of the reaction. The universal gas constant and temperature of the reaction are already given. of moles of reactantsĪpplying the above formula, we find Δn is 1. Where, “R” represents the universal gas constant “T” refers to the temperature at which the equilibrium is maintained and Δn is the difference in the number of moles of products and reactants. ![]() We know that the K p and Kc relation is K p = K c (RT) \(\bigtriangleup \)n Hence, K c can be represented as 5.12 x 10 -3 ![]() At equilibrium, rate of forward reaction is equal to rate of backward reaction.Similarly if k b is the rate constant for backward reaction then, the rate of backward reaction = k bCD.Thus the rate of forward reaction = k f AB, where k f is the velocity constant for the forward reaction.= 0.40 mol / 1dm 3 = 0.Consider a generalised reaction as follows:įrom the law of mass action, the rate at which A and B react is directly proportional to A and B. N = (concentration) x (volume) -> (concentration) = n / (volume) ): Now we convert our equilibrium amounts into concentrations (using We can now plug these numbers into our table: To calculate equilibrium amount from here we simply add on this change to the initial amounts: Also because they are our reactants, we must be losing amounts of them, so their changes in amount will be negative. This means that H 2 and I 2 amounts must change by half that. Volume of 1 dm 3.Ĭhange/mol 0.40 (we know that we have gained 0.40mol HI)Įquilibrium/mol 0.40 (at this point we only know equilibrium amount of HI)įrom here we know that HI has changed by 0.40 mol and also that there are two moles of HI in the overall equation. Starting amounts are: H 2 = 0.35 mol, I 2 = 0.25 mol and equilibrium amount of HI = 0.40 mol. Note: you must convert starting concentration to amounts (in moles) using container volume before putting them into the table. If the question gives you starting concentrations and equilibrium concentrations of one component rather than equilibrium concentrations of all components, you can calculate equilibrium concentrations using an ICE table (initial amount/mol, change in amount/mol, equilibrium amount/mol. For example for H 2 (g) I 2 (g) ⇄ 2HI(g), equilibrium concentrations are: To do the calculation you simply plug in the equilibrium concentrations into your expression for Kc. Note: the information must specify equilibrium concentration and not starting concentration. If you the information in the question gives you the equilibrium concentration then this step is very easy. ![]() It may be easier to understand this if you think of mol dm -3 as 'x' and so instead of replacing the molecules in the square brackets with mol dm -3, you replace them with x: Therefore in this case, K c has no units. You can see that the numerator, (mol dm -3) 2, is the same as the denominator, (mol dm -3 ) (mol dm -3 ) and so all the units cancel. The first step is to replace the molecules in the square brackets with the units of concentration. You can think of the units of concentration, mol dm -3, as 'x'. The method for calculating units is similar to very simple algebra (though don't be put off if you never got on with this in school, it really is very simple). ![]() Note: it's the concentration of the products over reactants, not the reactants over products. The concentration of each product raised to the power of its stoichiometric coefficient, divided by the concentration of each reactant raised to the power of its stoichiometric coefficient. T he equilibrium coefficient is given by: Where A and B are reactants, C and D are products and a, b, c and d are stoichiometric coefficients (relative number of moles of each molecule).
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