1
$\begingroup$

Solid $\ce{NaHCO3}$ is heated to $90~^\circ\mathrm{C}$. At equilibrium the total pressure of the gases produced is $0.545~\mathrm{atm}$. Calculate $\Delta G^\circ$ at $90~^\circ\mathrm{C}$ for the reaction.

$$\ce{2 NaHCO3(s) <=> Na2CO3(s) + H2O(g) + CO2(g)}$$

Using the $\Delta G^\circ = -\mathcal{R}T\cdot\ln(K)$ formula, I did the work below:

$$\Delta G^\circ = -(8.31)(273.15 + 90)\cdot\ln(0.545)$$

This gave me $1.83~\mathrm{kJ/mol}$ and therefore $3.66~\mathrm{kJ}$ for the entire reaction, $2~\mathrm{mol}$. Unfortunately, the answer is marked as $7.85~\mathrm{kJ}$, suggesting that $K=1$, why is this?

Shouldn't $K=(0.545)/1$ because of the gases present?

$\endgroup$
3
$\begingroup$

For your reaction:

$$\ce{2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g)}$$

$$K=P_{\ce{CO2}}\cdot P_{\ce{H2O}}$$

However, $K\ne0.545\text{ atm}$ because the sum of the pressures is $0.545\text{ atm}$ and the equilibrium constant is the product of the pressures.

$$P_{\ce{CO2}}+P_{\ce{H2O}}=0.545\text{ atm}$$

Since you started with no gas, and the stoichiometric coefficients of both gases are equal, then

$$P_{\ce{CO2}}=P_{\ce{H2O}}=\dfrac{0.545\text{ atm}}{2}$$

What is the value of $K$ now?

$\endgroup$
  • 1
    $\begingroup$ $K=\frac{{\left(\frac{0.545}{2}\right)}^2}{1}$ which means ∆G° is 7846J or 7.85kJ. Thank you for pointing out my mistake! $\endgroup$ – twrightsman May 18 '14 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.