Question

For the reaction:
$C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
$\Delta E=-1415KJ$. The $\Delta H$ at $27C$ is:

• A. $-1410KJ$
• B. $-1420KJ$
• C. $+1420KJ$
• D. $+1410KJ$

Answer 1

The correct option is B $-1420KJ$

Solution:- (B) $-1420kJ$

Given:-

$\Delta E=-1415kJ$

$T=27℃=(27+273)=200K$

For the given reaction-

${C_2{H}_4}_{\left(g\right)}+3{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

$\Delta n_g=n_P-n_R=2-(3+1)=-2$

Now as we know that,

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

$\Rightarrow \Delta H=-1415+(-2\times 8.314\times {10}^{-3}\times 300)$

$\Rightarrow \Delta H=-1415-5=-1420kJ$

Hence the value of $\Delta H$ is $-1420kJ$.