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.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied