• # question_answer For a gaseous reaction: $2{{A}_{2}}(g)+5{{B}_{2}}(g)\to 2{{A}_{2}}{{B}_{5}}(g)$ at ${{27}^{o}}C$ the heat change at constant pressure is found to be $-\,50160\,J$. Calculate the value of internal energy change $(\Delta E)$. Given that R = 8.314 J/K mol. A) - 34689 J B) - 37689 J C) - 27689 J D) - 38689 J

 [b] $2{{A}_{2}}\left( g \right)+5{{B}_{2}}\left( g \right)g\to 2{{A}_{2}}{{B}_{5}}\left( g \right);\Delta H=-50160J$$\Delta n=2-\left( 5+2 \right)=-5\,mol$ $\Delta H=\Delta E+{{\left( \Delta n \right)}_{g}}RT$ $-50160=\Delta E+{{\left( \Delta n \right)}_{g}}RT$ $\Delta E=-50160-\left( -5 \right)\left( 8.314 \right)\left( 300 \right)$ $=-\,50160+12471=-\,37689J$