logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

For a gaseous reaction $2A_2(g) + 5B_2(g) \rightarrow 2A_2B_5(g)$ at $ 27^{\circ}C$ the heat change at constant pressure is found to be $-50160\: J$. Calculate the value of internal energy change $(\Delta U). (R = 8.314 \: J/K\: mol)$

$\begin {array} {1 1} (a)\;37689\: J & \quad (b)\;-37689 \: J \\ (c)\;62631 \: J & \quad (d)\;-62631 \: J \end {array}$

Can you answer this question?
 
 

1 Answer

0 votes
The given chemical reaction, $2A_2(g) + 5B_2(g) \rightarrow 2A_2B_5(g); \: \Delta H = -50160\: J$
$\Delta n = 2 – (5+2) = -5\: mol$
Now applying the relation, $ \Delta H = \Delta U + (\Delta n)RT$
$-50160\: J = \Delta E + (\Delta n)RT$
$ \Rightarrow \Delta E = -50160 - \Delta nRT = -50160\: J – (-5\: mol)(8.314 \: J/K\: mol)(300\: K)$
$= -50160\: J + 12471\: J = -37689\: J$
Ans : (b)
answered Mar 13, 2014 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...