MCQ 11 Mark
On the basis of thermochemical equations (A), (B) and (C), find out which of the algebraic relationships given in options (i) to (iv) is correct:
A. C(graphite) $+ O _2(g) \rightarrow CO _2$ (g); $\triangle_r H= x KJ mol ^{-1}$
B. C(graphite) $+\frac{1}{2} O_2(g) \rightarrow C O(g) ; \Delta_r H= y KJ mol ^{-1}$
C. $C O(g)+\frac{1}{2} O_2(g) \rightarrow C O_2(g) ; \Delta_r H=z k J m o l=1$
A. C(graphite) $+ O _2(g) \rightarrow CO _2$ (g); $\triangle_r H= x KJ mol ^{-1}$
B. C(graphite) $+\frac{1}{2} O_2(g) \rightarrow C O(g) ; \Delta_r H= y KJ mol ^{-1}$
C. $C O(g)+\frac{1}{2} O_2(g) \rightarrow C O_2(g) ; \Delta_r H=z k J m o l=1$
- ✓$x=y+z$
- B$y=2 z-x$
- C$z=x+y$
- D$x=y-z$
Answer
View full question & answer→Correct option: A.
$x=y+z$
(a) $x=y+z$
Explanation: We have, C (graphite) $+ O _2(g) \rightarrow CO _2(g) ; \triangle_r H= x KJ mol ^{-1} \ldots .(1)$
C (graphite) $+\frac{1}{2} O _2(g) \rightarrow CO ( g ) ; \Delta_r H=y kJmol ^{-1} \ldots(2)$
Subtacting (1) and (2), we get;
$
\begin{aligned}
& CO(g)+\frac{1}{2} O_2(g) \rightarrow CO_2(g) ; \triangle_r H=(x-y) kJmol^{-1} \\
& \therefore z=x-y \Rightarrow x=y+z
\end{aligned}
$
Explanation: We have, C (graphite) $+ O _2(g) \rightarrow CO _2(g) ; \triangle_r H= x KJ mol ^{-1} \ldots .(1)$
C (graphite) $+\frac{1}{2} O _2(g) \rightarrow CO ( g ) ; \Delta_r H=y kJmol ^{-1} \ldots(2)$
Subtacting (1) and (2), we get;
$
\begin{aligned}
& CO(g)+\frac{1}{2} O_2(g) \rightarrow CO_2(g) ; \triangle_r H=(x-y) kJmol^{-1} \\
& \therefore z=x-y \Rightarrow x=y+z
\end{aligned}
$
