Question

From the following reactions at $298\, K$,
$CaC _{2}(s)+ 2 H _{2} O (l) \longrightarrow Ca ( OH )_{2}(s) + C _{2} H _{2}(g) DH ^{\circ}\left( kJ \,mol ^{-1}\right)-127.9 $
(B) $Ca (s)+\frac{1}{2} O _{2}(g) \longrightarrow CaO (s)-635.1$
(C) $CaO ( s )+ H _{2} O (I) \longrightarrow Ca ( OH )_{2}(s)-65.2$
(D) $C ( s )+ O _{2}( s ) \longrightarrow CO _{2}(s)-393.5$
(E) $C _{2} H _{2}(g)+\frac{5}{2} O _{2}(g) \longrightarrow 2 CO _{2}(g)+ H _{2} O (e)-1299.58$
Calculate the heat of formation of $CaC _{2}(s)$ at $298\, K$.

• A. $-59.82\, KJ \,mol ^{-1}$
• B. $+59.82\, KJ\,mol ^{-1}$
• C. $-190.22\, KJ \,mol ^{-1}$
• D. $+190.22 \,KJ \,mol ^{-1}$

Answer 1

Answer: (A)

We have to calculate $\Delta H^{o}$ for the following reaction in which one mole of $C a C_{2}(S)$ is formed form its elements.
$Ca (S)+2 C(S) \rightarrow C a C_{2}(S)$
This is obtained when we operate.
$(B)+(C)+z(D)-(E)-a$
$ \therefore \Delta H=(-635.1)+(-65.2)+2(-393.5)$
$-(-1299.58)-(-127.9) \,kJ \,m o l^{-1}$
$=-59.82\, kJ \,mol ^{-1}$