At what temperature is the following reaction feasible: Al2O3(s) + 3C(s) -> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol–1 Entropy (S) = +614 J K–1 mol–1 A. 2096.1 K B. 1273.8 K C. 477.1 K D. 1901.0 K

Option A. 2096.1 K

Explanation:

The following data were obtained from the question:

Al2O3(s) + 3C(s) —> 2Al(s) + 3CO(g)? Enthalpy (H) = +1287 kJ mol¯¹

Entropy (S) = +614 JK¯¹ mol¯¹

Temperature (T) =...?

Entropy, enthalphy and temperature are related by the following equation:

Change in Entropy (ΔS) = Change in Enthalphy (ΔH) /Temperature (T)

ΔS = ΔH/T

With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:

Enthalpy (H) = +1287 kJ mol¯¹ = 1287000 Jmol¯¹

Entropy (S) = +614 JK¯¹mol¯¹

Temperature (T) =...?

ΔS = ΔH/T

614 = 1287000/ T

Cross multiply

614 x T = 1287000

Divide both side by 614

T = 1287000/614

T = 2096.1 K

Therefore, the temperature at which the reaction will be feasible is 2096.1 K.

question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)