The $K_\mathrm{p}$ value of a reaction is $\pu{10 atm}$ at $\pu{300K}$ and $\pu{4atm}$ at $\pu{400K}$. 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

d. The difference between heat of reaction at constant pressure and that at constant volume is $RT$.

The correct answer given is option b.

But 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 need to use the van't Hoff equation? I tried using the ideal gas equation also but didn't get this $RT$. Also I think since change in enthalpy is negative so the reaction is exothermic.

  • 2
    $\begingroup$ What do you mean by "ko value" and "change in enthalpy is -ve"? Maybe this would be clearer if you used the math formatting this page provides. What are the reasons for you to "feel that this reaction is endothermic" if you haven't specified what reaction you are talking about? $\endgroup$
    – Philipp
    Aug 27, 2014 at 20:33
  • $\begingroup$ Sorry, but your question is very hard to understand. You clarified the "ko value" issue but the rest of my first comment remains. What reaction are you talking about and where does your claim that "the difference between the heat of reaction at constant pressure and that at constant volume is RT" come from? $\endgroup$
    – Philipp
    Aug 28, 2014 at 16:17
  • $\begingroup$ @Philipp please read the question again. This time I have give the actual question also. The correct answer to the actual question is the b part(which means that it is an incorrect statement) $\endgroup$
    – geek101
    Aug 29, 2014 at 15:44
  • $\begingroup$ Kp decrease with temperture so change in enthalpy is -ve so reaction is exothermic. $\endgroup$
    – geek101
    Aug 29, 2014 at 15:47
  • $\begingroup$ Again sorry, but your question still is very imprecise. I'm sure that it is not true for every possible reaction that the difference between heat of reaction at constant pressure and that at constant volume is RT. It might be true for some reaction involving ideal gases but the way the question is asked it is difficult for me to answer it. $\endgroup$
    – Philipp
    Aug 29, 2014 at 19:40

1 Answer 1


So the question is: Why is the difference between the heat of reaction at constant pressure and that at constant volume equal to $RT$?

First Part: heat of reaction

From the first law of thermodynamics we know

\begin{equation} \mathrm{d}U = \delta Q + \delta W = \delta Q - p \mathrm{d} V \end{equation}

where $U$ is the inner energy, $Q$ is the heat, $W$ is the work, $p$ is the pressure and $V$ is the volume. So, for a reaction at constant volume you get

\begin{equation} \mathrm{d}U = \mathrm{d} Q \ . \end{equation}

Now, for a reaction at constant pressure you can invoke the relationship

\begin{align} \mathrm{d}H &= \mathrm{d}U + p \mathrm{d}V + V \mathrm{d}p \\ &= \delta Q - p \mathrm{d} V + p \mathrm{d}V + V \mathrm{d}p \\ &= \delta Q + V \mathrm{d}p \\ &\overset{\mathrm{d}p = 0}{=} \mathrm{d} Q \end{align}

where $H$ is the enthalpy. This gives you the relationships that describe the change in heats released during reactions at constant volume and constant pressure. The so called heat of reaction is defined as

\begin{equation} \Delta Q = \biggl( \frac{\partial Q}{\partial \xi} \biggr) \end{equation}

where $\xi$ is the extent of reaction.

Second Part: temperature dependence

Now using the equality of mixed partials and the definitions of the heat capacities at constant pressure $C_{p} = \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p}$ and constant volume $C_{V} = \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V}$ you get for constant volume ($\mathrm{d}U = \mathrm{d} Q$):

\begin{align} \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta U}{\partial T} \Biggr)_{V} &= \left( \frac{ \Bigl( \frac{\partial U}{\partial \xi} \Bigr)_{T, V}}{\partial T} \right)_{V} = \Biggl( \frac{\partial^{2} U}{\partial \xi \partial T} \Biggr)_{V} \\ &= \Biggl( \frac{\partial^{2} U}{\partial T \partial \xi} \Biggr)_{V} = \left( \frac{ \Bigl( \frac{\partial U}{\partial T} \Bigr)_{V} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial C_{V}}{\partial \xi} \Biggr)_{T, V} = \Delta C_{V} \end{align}

and for constant pressure ($\mathrm{d}H = \mathrm{d} Q$):

\begin{align} \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} = \Biggl(\frac{\partial \Delta H}{\partial T} \Biggr)_{p} &= \left( \frac{ \Bigl( \frac{\partial H}{\partial \xi} \Bigr)_{T, p}}{\partial T} \right)_{p} = \Biggl( \frac{\partial^{2} H}{\partial \xi \partial T} \Biggr)_{p} \\ &= \Biggl( \frac{\partial^{2} H}{\partial T \partial \xi} \Biggr)_{p} = \left( \frac{ \Bigl( \frac{\partial H}{\partial T} \Bigr)_{p} }{\partial \xi} \right)_{T} = \Biggl(\frac{\partial C_{p}}{\partial \xi} \Biggr)_{T, p} = \Delta C_{p} \end{align}

where $T$ is the temperature and $\Delta C_{i}$ is change in the molar heat capacities of the reactants and products during the reaction at constant $i$ (where $i$ is either $p$ or $V$):

\begin{align} \Delta C_{i} = \sum_{k} \nu_{k} c_{i_{k}} \end{align}

where $\nu_{k}$ and $c_{i_{k}}$ are the stochiometric coefficient and the molar heat capacity at constant $i$ of the $k^{\text{th}}$ component in the reaction, respectively.

When you integrate the equations for the temperature dependences of the reaction heats you get:

\begin{align} \text{constant volume: } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{V} \, \mathrm{d} T = \left( \Delta Q \right)_{V} = \int \Delta C_{V} \, \mathrm{d} T \\ \text{constant pressure: } &\int \biggl(\frac{\partial \Delta Q}{\partial T} \biggr)_{p} \, \mathrm{d} T = \left( \Delta Q \right)_{p} = \int \Delta C_{p} \, \mathrm{d} T \ . \end{align}

So, the difference of the reaction heats at constant volume and constant pressure would be:

\begin{align} \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T \ . \end{align}

Third Part: ideal gases

Now, why would this boil down to $RT$? This works when you consider reactions that involve only ideal gases. For ideal gases you have (see here):

\begin{align} c_{p} - c_{V} = R \ . \end{align}

So, the integral simplifies to

\begin{align} \int \left( \Delta C_{p} - \Delta C_{V} \right) \, \mathrm{d} T &= \int \sum_{j} \nu_{j} \bigl( \underbrace{ c_{p_{j}} - c_{V_{j}} }_{= \, R} \bigr) \, \mathrm{d} T \\ &= R \int \sum_{j} \nu_{j} \, \mathrm{d} T \\ &= R \sum_{j} \nu_{j} \int \mathrm{d} T \\ &= R T \sum_{j} \nu_{j} \end{align}

where $\sum_{j} \nu_{j}$ will be some rational number but not necessarily equal to 1 - it depends on the reaction. This is basically the result you were looking for although the reaction heat difference will only be proportional to $RT$ with the proportionality constant being $\sum_{j} \nu_{j}$:

\begin{align} \left( \Delta Q \right)_{p} - \left( \Delta Q \right)_{V} = RT \sum_{j} \nu_{j} \ . \end{align}


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