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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?

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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?

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    $\begingroup$ $K=\frac{{\left(\frac{0.545}{2}\right)}^2}{1}$ which means ∆G° is 7846J or 7.85kJ. Thank you for pointing out my mistake! $\endgroup$ May 18, 2014 at 19:23

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