Consider the reactions C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H = −758 kJ mol−1 2C(s) + 2H2(g) → C2H4(g) ∆H = +52 kJ mol−1 H2(g) + O2(g) → H2O(g) ∆H = −242 kJ mol−1 The enthalpy of formation of carbon monoxide is A −111 kJ mol−1 B −163 kJ mol−1 C −222 kJ mol−1 D -464 kJ mol−1

C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g)       ∆H1 = −758 kJ mol−1       1)

2C(s) + 2H2(g) → C2H4(g)                           ∆H2 = +52 kJ mol−1        2)

H2(g) + O2(g) → H2O(g)                               ∆H3 = −242 kJ mol−1     3)

Now, enthalpy of formation of carbon monoxide is given by :

∆H = ∆H1 + ∆H2 - ∆H3

∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1  

∆H = −464 kJ mol−1

Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.

Hence, this is the required solution.

describe matter a earth

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.