Consider the reaction X2(g) ⟶2 X(g) . When a vessel initially containing 755 torr of X2 comes to

Consider the reaction X2(g) ⟶2 X(g) . When a vessel...

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations or partial pressures of reactants and products remain constant over time. Understanding chemical equilibrium allows one to relate changes in conditions, such as temperature or pressure, to shifts in the equilibrium position.

Equilibrium Constant (Kp)

The equilibrium constant, expressed in terms of partial pressures (Kp) for gaseous reactions, quantifies the ratio of the product pressures to reactant pressures at equilibrium, each raised to the power of their stoichiometric coefficients. It provides a measure of the extent of a reaction at a given temperature, and changes in Kp with temperature can offer insights into the thermodynamic properties of the reaction.

Van't Hoff Equation

The van't Hoff equation describes the temperature dependence of the equilibrium constant. It connects the change in the natural logarithm of the equilibrium constant to the reciprocal of the temperature and the standard enthalpy change of the reaction. This relationship is pivotal in determining the reaction enthalpy (?H°) from experimental equilibrium data at different temperatures.

Standard Reaction Enthalpy (?H°)

The standard reaction enthalpy (?H°) represents the heat absorbed or released during a reaction carried out under standard conditions. It is a key thermodynamic parameter indicating whether a reaction is endothermic or exothermic. ?H° can be derived from the temperature dependence of the equilibrium constant using the van't Hoff equation, linking measurable equilibrium pressures with the heat change of the reaction.

Transcript

00:01 In this question we've been given two states, that is the initial state and the final state, and we've been asked to determine the standard end of the change associated with that change from state, from initial conditions up to final conditions.

00:15 So when we are given this type of information, the one equation that comes to mind is lynn k2 over k1 is equal to the standard enthalpy change or the reaction divided by r, this is negative, then we have 1 over t2 minus 1 over t 1 .1 so all we want to do here is to calculate these standard end up change so the main task here is to be determining the unknowns here in this equation so that we plug them in here for us to then make this standard enthalpy change the subject of the formula so the first thing that we're going to do is to determine our k2 and we know that k2 k2 the k2 the k here is the equilibrium constant which relates the partial pressure of the products to that of the reactants here we've got x2 being converted into x2 2 2 moles of x so the kp expression here is going to be the partial pressure of x squared divided by the partial pressure of x2 so remember this is an equilibrium constant so these are not just the partial pressures, but the partial pressures at equilibrium.

01:35 So the main task here will be to determine these partial pressures at equilibrium.

01:40 This we are going to do for k1 and our k2, that we are then going to substitute into this formula for us to then determine the standard end of the change of that reaction.

01:53 So first of all, we have to have an ice table, or we have x2, and two moles of.

02:00 Of x remember we have the initial and we have a certain change and we use those two we add those to determine the pressure at equilibrium and if we are to look at this we've been given toes so we always have to make it a habit to convert these into into atmospheres so for example px2 this is going to be equal to remember this is the initial the initial pressure px2 not this is 752.

02:35 And for us to convert to atmosphere we divide by 760, then this gives us 0 .99342 in atmosphere.

02:46 And px, px, p at equilibrium, this is going to be equal to 103, 103, 2, divided by 760.

03:02 So this is going to be 0 .1 -3553 and this is also in atmosphere.

03:09 So if we have got an initial pressure here, this is going to be 0 .99342 atmosphere.

03:17 And before this equation starts, that is under initial conditions, there won't be any product, so this is going to be equal to 0.

03:25 So if we are to look at this, let's say this changes by a factor of x.

03:29 We're going to say minus x this negative sign shows the partial pressure is decreasing with time because this is a reactant so if we look at this we have x2 that is forming 2x so if this changes if x2 changes by a factor of x 1 mole of x2 is forming 2 moles of x so if x changes by a factor of x it means this is going to change by a factor of 2x because we are looking at this documentary coefficients 1 is to 2 so this is going to be a positive so at equilibrium we're going to have 0 .99 3 4 2 minus x and at equilibrium we're going to have the partial pressure of x is going to be 2x but we know the pressure of of x at equilibrium this is what we determined here so what we are saying here is our 2x is equal to 0 .13553.

04:33 So the value of x here is going to be equal to 0 .067765.

04:40 So the partial pressure of of x2 at equilibrium, this is going to be equal to, we determine this to be this expression right here.

04:53 So this is going to be equal to 0 .0 .93345.

04:59 Minus x and this is equal to 0 .99 -345 minus 0 .0 .0 .67765.

05:10 So the partial pressure of x2 at equilibrium, this is going to be equal to 0 .13553 and this is in atmosphere.

05:20 So we now have the partial pressures at equilibrium.

05:26 We can determine our k1.

05:28 Remember our k1, this is a equal to the partial pressure of x raised to the power of two divided by the partial pressure of x2 at equilibrium and this is going to be equal to 0 .13553 and we have to square this divided by the partial pressure of x2 at equilibrium and this is going to be equal to 0 .01 0.

06:04 This is 0 .9 .0.

06:05 This is 0 .9.

06:07 0 .92.

06:11 This is 0 .92555.

06:16 Here, if we make this subtraction here, this is going to be equal to 0 .925655 in atmosphere.

06:32 So moving forward, we now have our k1, k1 being equal to 0 .019844.

06:49 Now that we have our k1, we then move on to determine our k2...

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