[Jai]

If supposing the kp value of a reaction is 10 Atm at 300K and 4atm at 400K. Then how can we say that the difference between the heat of reaction at constant pressure and that at constant volume is RT for the above reaction? How do we go about approaching this problem? Do we use the vant Hoffs equation? I tried the ideal gas equation also but didn't get this RT. Also I think since change in enthalpy is -ve so reaction is exothermic. But how do we go ahead solving this problem? please help me. (The exact question is - the kp value of a reaction is 10 Atm at 300K and 4atm at 400K. The incorrect statement about the reaction is- A. The reaction is exothermic B. The rate of forward direction is more than that of backward reaction. C. The rate of backward reaction increases more than that of forward reaction with increase of temperature(exoteric I think) D. The difference between heat of reaction at constant pressure and that at constant volume is RT the answer is b(meaning that is is incorrect) but I can't understand why the last part is true then?.)

[Corribus]

Well, the nice thing is that you can arrive at the correct answer (D) simply by process of elimination, without actually knowing whether D is incorrect, or how to show that it is:

Increasing the temperature of the reaction favors formation of products when the reaction is endergonic, and favors the formation of reactants when the reaction is exergonic. This is somewhat counterinuitive to a lot of students, so if you have trouble understanding why this is the case, you can read more about it here:

http://www.chemicalforums.com/index.php?topic=68299.msg246224#msg246224

A less rigorous but perhaps easier-to-remember explanation is that an equilibrium will respond to a change in such a way as to counteract that change. So, if you heat a reaction up by increasing the temperature, it will respond by "removing heat" - that is, it will favor an endothermic product. If the overall equilibrium is endothermic, this means formation of products; if the overall equilibrium is exothermic, this means formation of reactant.

In your problem you are increasing the temperature and lowering the equilibrium constant, which means you are favoring reactants by increasing the temperature. This is because Kp for a simple reaction A --> B is the partial pressure of the product divided by the partial pressure of the reactant. Thus if your Kp is decreasing when you increase the temperature, it means that the amount of B, relative to A, is going down. Increasing the temperature decreases the equilibrium constant, which means the process is exothermic. Therefore A can't be incorrect.

We'll get back to B in a moment, although we get a hint of why it's incorrect from choice C.

C follows from the discussion above. If the reaction is exothermic, and the equilibrium constant decreases as temperature increases, this means the reactants are favored as the temperature increases. If the reactants are favored, it means the reverse direction of the reaction is favored. Therefore C can't be incorrect.

D isn't incorrect because it basically follows from the ideal gas law and the first law of thermodynamics. Under constant volume, the heat given off (or absorbed) during a process is equal to the change in internal energy. At constant pressure, the heat given off (or absorbed) is the enthalpy change - since most chemical reactions are done at constant pressure, the enthalpy of reaction is often called the heat of reaction, even though they are technically different things. The difference between the heat change at constant pressure (ΔH) and constant volume (ΔU) is the additional expansion work done, Δ(PV). In the ideal gas approximation, this is equal to Δ(n)RT. The quantity is usally pretty small in condensed phases, such that the constant volume and constant pressure heats of reaction are nearly identical. In gas phase reactions, where the pressure and volume changes can be large, it is not negligible.

This leaves (B) as the only incorrect answer. And it is incorrect because at equilibrium the rates of the forward and backward reactions are identical (at any given temperature).

[mjc123]

To be a bit more specific about D: Look at the units of Kp. (Though admittedly people are often sloppy about the units of equilibrium constants - and rate constants.)
For a general gaseous reaction aA + bB  cC + dD
Kp = P_A^a.P_B^b/P_C^c.P_D^d
The dimensions of Kp are (pressure)^a+b-c-d = (pressure)^-Δn where Δn is the change in the number of moles of gas.
As the units of Kp in your example are atm, you can tell that Δn = -1, and ΔH = ΔU - RT (and not e.g. ΔU + 2RT).

[mjc123]

Sorry, that should of course be

Kp = P_C^c.P_D^d/P_A^a.P_B^b
The dimensions of Kp are (pressure)^c+d-b-a = (pressure)^Δn where Δn is the change in the number of moles of gas.
As the units of Kp in your example are atm, you can tell that Δn = +1, and ΔH = ΔU + RT (and not e.g. ΔU + 2RT).

Talk about people being sloppy....It's Monday morning!!