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.

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g