Question

If the heat of combustion of carbon monoxide at constant volume and at ${17}^o$C is $-283.3$ kJ, then its enthalpy of combustion at constant pressure($R=8.314Jdegre{e}^{-1}mo{l}^{-}$)

• A. $-284.5$ kJ
• B. $284.5$ kJ
• C. $384.5$ kJ
• D. $-384.5$ kJ

Answer 1

The correct option is A $-284.5$ kJ

Solution:- (A) $-284.5kJ$

Heat change at constant volume for the combustion of carbon monoxide $=-283.3kJ$

${CO}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{C{O}_2}_{\left(g\right)}$

From the above reaction,

$\Delta n_g=n_P-n_R=1-(\frac{1}{2}+1)=-\frac{1}{2}$

Temperature $\left(T\right)=17℃=(17+273)K=290K\left(\text{Given}\right)$

Now from first law of thermodynamics,

$\Delta H=\Delta E+\Delta n_{g}RT$

$\Delta H=-283.3+(-\frac{1}{2})\times 8.314\times {10}^{-3}\times 290$

$\Rightarrow \Delta H=-283.3-1.205=-284.505kJ$

Hence the heat of reaction at constant pressure will be $-284.5kJ$.