Question

Half a litre of $CO _{2}$ is passed over red-hot carbon. The volume becomes $700 mL$. Find the composition of the product assuming that all gases are measured at S.T.P.

Answer 1

From equation we see that 1 vol. of $CO _{2}$ on complete conversion to $CO$ produces 2 vol. of $CO$. Had all the $CO _{2}$, i.e., $500 mL$ taken part in the reaction, the vol. of $CO$ formed would have been $(2 \times 500) mL =1000 mL$. But the vol. of product is $700 mL$. It is thus clear that a portion of $CO _{2}$ has changed to $CO$. Thus the mixture having a vol. of $700 mL$ contains both unchanged $CO _{2}$ and $CO$ formed.

Let $x mL$ be the vol of $CO _{2}$ that changed to $CO$. Therefore, $(500-x) mL$ is the vol. of $CO _{2}$ that is not reduced when $CO _{2}$ is passed over red-hot carbon; and $2 x mL$ is the vol. of $CO$ formed. Now, vol. of $CO$ formed + vol. of unchanged $CO _{2}=700 mL$

or, $2 x+(500-x)=700 mL \quad$ or, $x=200 mL$

Hence, the ultimate product is a mixture of

$(500-200) mL =300 mL CO _{2} \text {, and }(200 \times 2) mL =400 mL CO \text {. }$