Question

what is $\delta g^\circ_{\text{rxn}}$ in kj/mol for this reaction at 417 $\circ$c? please report an integer, without units.\
$\ce{4 hno3(g) + 5 n2h4(l) -> 7 n2(g) + 12 h2o(l)}$\
$\delta h^\circ_{\text{rxn}} = -3147$ kj/mol\
$\delta s^\circ_{\text{rxn}} = 507.6$ j/mol k

Explanation:

Step1: Convert temperature to Kelvin

The temperature is given in Celsius, so we convert it to Kelvin using the formula \( T(K) = T(^\circ C) + 273.15 \).
\( T = 417 + 273.15 = 690.15 \, K \)

Step2: Convert entropy to kJ/(mol·K)

The entropy change \( \Delta S^\circ_{rxn} \) is given in J/(mol·K), so we convert it to kJ/(mol·K) by dividing by 1000.
\( \Delta S^\circ_{rxn} = \frac{507.6 \, J/(mol·K)}{1000} = 0.5076 \, kJ/(mol·K) \)

Step3: Use the Gibbs free energy formula

The formula for Gibbs free energy change is \( \Delta G^\circ_{rxn} = \Delta H^\circ_{rxn} - T\Delta S^\circ_{rxn} \).
Substitute the values: \( \Delta H^\circ_{rxn} = -3147 \, kJ/mol \), \( T = 690.15 \, K \), and \( \Delta S^\circ_{rxn} = 0.5076 \, kJ/(mol·K) \)
\( \Delta G^\circ_{rxn} = -3147 - (690.15 \times 0.5076) \)
First, calculate \( 690.15 \times 0.5076 \approx 350.3 \)
Then, \( \Delta G^\circ_{rxn} = -3147 - 350.3 = -3497.3 \approx -3497 \)

Answer:

-3497