Consider the reaction
2SO₂(g) + O₂(g) 2SO₃(g)
For which ΔH° = -200. kJ and ΔS° = -187 J/K at 25°C. Assuming that ΔH° and ΔS° are independent of temperature, calculate the temperature where K_p = 1.

Question

Consider the reaction
2SO₂(g) + O₂(g) Consider the reaction 2SO2(g) + O2(g) 2SO3(g) For which ΔH° = -200. kJ and ΔS° = -187 J/K at 25°C. Assuming that ΔH° and ΔS° are independent of temperature, calculate the temperature where Kp = 1. A) 200. K B) 1070 K C) 2070 K D) 970. K E) none of these

Consider the reaction 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) 2SO<sub>3</sub>(g) For which ΔH° = -200. kJ and ΔS° = -187 J/K at 25°C. Assuming that ΔH° and ΔS° are independent of temperature, calculate the temperature where K<sub>p</sub> = 1. A) 200. K B) 1070 K C) 2070 K D) 970. K E) none of these

2SO₃(g)
For which ΔH° = -200. kJ and ΔS° = -187 J/K at 25°C. Assuming that ΔH° and ΔS° are independent of temperature, calculate the temperature where K_p = 1.

Quizplus · Accepted Answer

The equation to use is: ΔG° = ΔH° - TΔS° = -RTln(K) where R is the gas constant, T is the temperature in Kelvin, and ln(K) is the natural logarithm of the equilibrium constant. We want to find the temperature at which K = 1, so we can set K_p equal to 1 and solve for T: 1 = e^(-ΔG°/RT) ln(1) = -ΔG°/RT 0 = -ΔG°/RT T = -ΔG°/(Rln(K)) Plugging in the values given: T = -(-200000 J)/(8.314 J/K/mol * ln(1)) T = 1070 K Therefore, the answer is B) 1070 K.

Consider the Reaction 2SO2(g) + O2(g) 2SO3(g)

Consider the Reaction 2SO₂(g) + O₂(g) 2SO₃(g)