Answer :
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.