Question

$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
$\delta E-\delta H$ for this reaction ${27}^0$C will be :

• A. $+1247.1$J
• B. $-1247.1$J
• C. $-6235.5$J
• D. $+6235.5$J

Answer 1

Answer: (D)

$+6235.5$J

Solution:- (D) $+6235.5J$

${C_2{H}_6}_{\left(g\right)}+\frac{7}{2}{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

$\Delta n_g$ for the above reaction-

$\Delta n_g=n_P-n_R=(2+0)-(1+\frac{7}{2})=-\frac{5}{2}$

$T=27℃=(27+273)=300K\left(\text{Given}\right)$

Now, from first law of thermodynamics,

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

$\Rightarrow \Delta E-\Delta H=-\Delta n_{g}RT=-[(-\frac{5}{2})\times 8.314\times 300]=+6235.5J$

Hence the value of $\Delta E-\Delta H$ for the given reaction is $+6235.5J$