| 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
Correct Answer: B
Solution :
| [b] \[2{{A}_{2}}\left( g \right)+5{{B}_{2}}\left( g \right)g\to 2{{A}_{2}}{{B}_{5}}\left( g \right);\Delta H=-50160J\]\[\Delta n=2-\left( 5+2 \right)=-5\,mol\] |
| \[\Delta H=\Delta E+{{\left( \Delta n \right)}_{g}}RT\] |
| \[-50160=\Delta E+{{\left( \Delta n \right)}_{g}}RT\] |
| \[\Delta E=-50160-\left( -5 \right)\left( 8.314 \right)\left( 300 \right)\] |
| \[=-\,50160+12471=-\,37689J\] |
You need to login to perform this action.
You will be redirected in
3 sec
