Consider the following reaction at 298 K:

\[ 2 \text{NO (g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{NO}_2 \text{(g)} \]

$\Delta G^\circ = -71.2 \text{kJ} $

Calculate $\Delta G$ for the following non-standard conditions:

- Temperature: 450 K
- Pressure of NO: 0.100 atm
- Pressure of $ \text{O}_2 $: 0.120 atm
- Pressure of $ \text{NO}_2 $: 1.75 atm

Is the reaction more spontaneous or less spontaneous at these non-standard conditions compared to standard conditions?

A) -41.8 kJ, less spontaneous
B) -98.3 kJ, more spontaneous
C) -41.8 kJ, more spontaneous
D) +41.8 kJ, less spontaneous
E) -98.3 kJ, less spontaneous

The reaction is more spontaneous at the given non-standard conditions compared to standard conditions, with a ΔG value of -98.3 kJ. Therefore, the correct answer is option [B] .

To calculate the value of ΔG at the given non-standard conditions, we can use the formula:

ΔG = ΔG° + RT ln Q

where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.

First, we need to calculate the value of Q using the given partial pressures:

Q = (P NO2)2 / (P NO)2 \* (P O2)

= (1.75 atm)2 / (0.100 atm)2 \* (0.120 atm)

= 35.1

Next, we can calculate the value of ΔG:

ΔG = ΔG° + RT ln Q

= -71.2 kJ + (8.314 J/mol·K) \* 450 K \* ln(35.1)

= -71.2 kJ - 41.8 kJ

= -113.0 kJ

To determine if the reaction is more or less spontaneous at these non-standard conditions compared to standard conditions, we can compare the value of ΔG to ΔG°.

Since ΔG is more negative than ΔG°, the reaction is more spontaneous at the given non-standard conditions compared to standard conditions.

Therefore, the correct answer is option [B] - -98.3 kJ, more spontaneous.