# Calculate DeltaS^Theta at 25^circC?

## The reaction ${\text{CH"_(4text[(g)]) + "N"_2""_text[(g)] -> "HCN"_text[(g)] + "NH"_3}}_{\textrm{\left(g\right)}}$ At ${25}^{\circ} C$ and $1 \text{atm}$ pressure has $\Delta {H}^{\Theta} = 164 k J$ $m o {l}^{-} 1$ and $\Delta {G}^{\Theta} = 159 k J$ $m o {l}^{-} 1$. Calculate $\Delta {S}^{\Theta}$ at ${25}^{\circ} C$. A) $400 J$ ${K}^{-} 1$ $m o {l}^{-} 1$ B) $100 J$ ${K}^{-} 1$ $m o {l}^{-} 1$ C) $2 J$ ${K}^{-} 1$ $m o {l}^{-} 1$ D) $17 J$ ${K}^{-} 1$ $m o {l}^{-} 1$ E) $70 J$ ${K}^{-} 1$ $m o {l}^{-} 1$

May 16, 2018

D. $17 J$` $m o {l}^{-} 1$

#### Explanation:

The equation for Gibbs free energy is given by:
$\Delta G = \Delta H - T \Delta S$

In this case $\Delta S = \frac{\Delta H - \Delta G}{T}$

$\Delta S = \frac{164000 - 159000}{25 + 273} = \frac{5000}{298} = 16.8 \approx 17 J$ ${K}^{-} 1$ $m o {l}^{-} 1 \equiv D$

May 16, 2018

$\text{D) 17 J/(K mol)}$

#### Explanation:

Use this equation

${\text{ΔG"^@ = "ΔH"^@ - "TΔS}}^{\circ}$

On rearranging

$\text{ΔS"^@ = ("ΔH"^@ - "ΔG"^@)/("T") = "164 kJ/mol – 159 kJ/mol"/"298 K" ≈ "17 J/(K mol)}$