Question

The difference in ∆h and ∆E for the combustion of methane at 25°c would be

Answer 1

Answer: 2476.38 J/mol

Explanation: We know ΔH=ΔEΔngRT

or ΔH−ΔE=ΔngRT

Delta H and Delta E is same for combustion of methane hence the difference is 0

CH4+2O2=CO2+2H2O

Here Δng=18−17=1

So, ΔH−ΔE=1×8.31J/k/mol×298K=2476.38 J/mol

Answer 2

The difference in ∆H and ∆E for the combustion of methane at 25°c would be $-2RT$

Explanation:

The reaction for combustion of reaction is

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$

$\Delta H=\Delta E+\Delta n_g\times RT$

where,

$\Delta H$ =  enthalpy of the reaction

$\Delta E$= internal energy of the reaction

$\Delta n_g$ = change in the moles of the reaction = Moles of product - Moles of reactant = 1 - 3 = -2

R = gas constant

T = temperature

Putting in the values we get:

$\Delta H=\Delta E+(-2)\times RT$

$\Delta H-\Delta E=(-2)\times RT$

Learn More about internal energy change and enthalpy change

https://brainly.in/question/12558547

https://brainly.in/question/12355019