Question

For the reaction $H_2(g)+I_2(g)\to 2HI(g)$ the change in enthalpy ($\Delta H$) will be

• A. $=\Delta E$
• B. $>\Delta E$
• C. $<\Delta E$
• D. either (b) or (c)

Answer 1

Answer: (A)

$H_2(g)+I_2(g)\to 2Hl(g)$

$\Delta n=$ no. of product speices - no. of reactant spccies

$=2-(1+1)=0$

As $\Delta H=\Delta E+\Delta nRT$

$\Rightarrow \Delta H=\Delta E+0\times RT$

$\Rightarrow \Delta H=\Delta E$