Question

given $\delta h = -46.2$ kj/mol and $\delta s = -389.0$ j·mol⁻¹·k⁻¹, determine $\delta g$ for the reaction below at $298.0$ °c.
\\ 3\\,\text{h}_2\\,\text{(g)} + \text{n}_2\\,\text{(g)} \
ightleftharpoons 2\\,\text{nh}_3\\,\text{(g)} \\

answer:
\boxed{} kj/mol

Explanation:

Step1: Convert temperature to Kelvin

The temperature is given in Celsius, so we convert it to Kelvin using the formula \( T(K) = T(^\circ C) + 273.15 \).
\( T = 298.0 + 273.15 = 571.15 \, K \)

Step2: Convert ΔS units to kJ/(mol·K)

Given \( \Delta S = -389.0 \, J \cdot mol^{-1} \cdot K^{-1} \), we convert joules to kilojoules by dividing by 1000.
\( \Delta S = \frac{-389.0}{1000} = -0.3890 \, kJ \cdot mol^{-1} \cdot K^{-1} \)

Step3: Use the Gibbs free energy formula \( \Delta G = \Delta H - T\Delta S \)

Substitute \( \Delta H = -46.2 \, kJ/mol \), \( T = 571.15 \, K \), and \( \Delta S = -0.3890 \, kJ \cdot mol^{-1} \cdot K^{-1} \) into the formula.
\( \Delta G = -46.2 - (571.15 \times -0.3890) \)
First, calculate \( 571.15 \times -0.3890 = -222.17735 \)
Then, \( \Delta G = -46.2 - (-222.17735) = -46.2 + 222.17735 = 175.97735 \approx 176.0 \, kJ/mol \) (Wait, this seems incorrect. Wait, no, wait: Wait, the formula is \( \Delta G = \Delta H - T\Delta S \). Wait, let's recheck the temperature. Wait, 298.0 °C? Wait, no, 298.0 K is 25 °C. Wait, maybe the problem has a typo? Wait, no, the user wrote 298.0 °C. Wait, but 298.0 °C is 571.15 K. But let's check again. Wait, maybe I made a mistake. Wait, the standard temperature is 298 K (25 °C). Wait, maybe the problem meant 298.0 K? Let's assume that maybe it's a typo, and it's 298.0 K (25 °C). Let's recalculate with T = 298.0 K.

Step3 (Corrected): Use T = 298.0 K (assuming a typo, since 298.0 °C is unusual, and 298 K is standard)

\( \Delta G = \Delta H - T\Delta S \)
\( \Delta H = -46.2 \, kJ/mol \)
\( T = 298.0 \, K \)
\( \Delta S = -389.0 \, J \cdot mol^{-1} \cdot K^{-1} = -0.3890 \, kJ \cdot mol^{-1} \cdot K^{-1} \)
\( \Delta G = -46.2 - (298.0 \times -0.3890) \)
Calculate \( 298.0 \times -0.3890 = -116.922 \)
Then, \( \Delta G = -46.2 - (-116.922) = -46.2 + 116.922 = 70.722 \approx 70.7 \, kJ/mol \)? Wait, no, that's not right. Wait, no, the formula is \( \Delta G = \Delta H - T\Delta S \). Wait, \( \Delta S \) is negative, so -TΔS is -T*(-|ΔS|) = T|ΔS|. So:

Wait, let's do it correctly. Let's use T = 298.0 K (since 298.0 °C is 571 K, but let's check the problem again. The user wrote 298.0 °C. Maybe it's correct. Let's proceed with T = 571.15 K.

\( \Delta G = -46.2 \, kJ/mol - (571.15 \, K \times -0.3890 \, kJ \cdot mol^{-1} \cdot K^{-1}) \)
\( = -46.2 + (571.15 \times 0.3890) \)
Calculate 571.15 0.3890: 571.15 0.3 = 171.345, 571.15 0.08 = 45.692, 571.15 0.009 = 5.14035. Sum: 171.345 + 45.692 = 217.037 + 5.14035 = 222.17735
So \( \Delta G = -46.2 + 222.17735 = 175.97735 \approx 176.0 \, kJ/mol \)

But this seems high. Wait, maybe the temperature is 298.0 K (25 °C). Let's check the original problem again. The user wrote "298.0 °C". Maybe it's a mistake. Let's confirm the formula. The Gibbs free energy change is given by \( \Delta G = \Delta H - T\Delta S \), where T is in Kelvin.

So, if T is 298.0 °C, that's 571.15 K. Let's proceed with that.

Answer:

\( 176.0 \) (or if T is 298 K, 70.7, but based on the given 298.0 °C, it's 176.0)