You can find this on google easily
Explanation:
a) when zinc burnt in oxygen.
2Zn + O2 -----∆-----> 2ZnO(black residue)
b) when carbon burnt in oxygen.
C+O2----∆---> CO2.
c) when sulphur burnt in oxygen.
S+O2-----∆-----> SO2.
d) when Calcium burnt in oxygen.
2Ca+O2-----∆-----> 2CaO(black residue)
e) when Magnesium burnt in oxygen.
2Mg+O2-----∆----> 2MgO.
f) when sodium burnt in oxygen.
4Na+O2----∆-----> 2Na2O.
hope all these reactions help you.
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature 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:
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
