Question

Equal volumes of $C_2{H}_2$ and $H_2$ are combusted under identical conditions. The ratio of heat evolved for $C_2{H}_2$ and $H_2$ is:
$H_2\left(g\right)+1/2O_2\left(g\right)\to H_{2}O\left(g\right),$ $\Delta H=-241.8$ kJ/mol
$C_2{H}_2\left(g\right)+5/2O_2\left(g\right)\to 2C{O}_2\left(g\right)+H_{2}O\left(g\right)$, $\Delta H=-1300$ kJ/mol

• A. $\frac{5.37}{1}$
• B. $\frac{1}{5.37}$
• C. $\frac{1}{1}$
• D. none of these

Answer 1

The correct option is A $\frac{5.37}{1}$

According to Avogadro's law: Equal volumes of all gases, at the same temperature and pressure, have the same number of molecules/moles

Thus equal volumes of $C_2{H}_2$ and $H_2$ under identical conditions have same number of moles say n.

Given that:

$H_2\left(g\right)+1/2O_2\left(g\right)\to H_{2}O\left(g\right),$ $\Delta H=-241.8$ kJ/mol

$C_2{H}_2\left(g\right)+5/2O_2\left(g\right)\to 2C{O}_2\left(g\right)+H_{2}O\left(g\right)$, $\Delta H=-1300$ kJ/mol

Thus, the ratio of heat evolved for $C_2{H}_2$ to $H_2$ = ratio of enthalpy of combustion of $C_2{H}_2$ to $H_2$

$\therefore \frac{q_{C_2{H}_2}}{q_{H_2}}=\frac{n.{\Delta }_{C_2{H}_2}H}{n.{\Delta }_{C_2{H}_2}H}$

$\frac{q_{C_2{H}_2}}{q_{H_2}}=\frac{-1300}{-241.8}$

or $\frac{q_{C_2{H}_2}}{q_{H_2}}=\frac{5.37}{1}$ for equal volume of the two under identical condition of pressure and temperature.