Question

on combustion of 1 gram of ch4 at constant pressure at the release amount of energy is used to is the temperature 1kg is 2 by 5 degree Celsius if the specific heat of H2O is one calorie program then enthalpy of combustion of ch4 will be​

Answer 1

Enthalpy can be determined by

$ΔH=ΔU+Δn </p><p>g</p><p> </p><p> RT$

Combustion of methane:-

$CH </p><p>4</p><p> </p><p> </p><p> </p><p> + </p><p>(g)</p><p>2O </p><p>2</p><p> </p><p> </p><p> </p><p> ⟶ </p><p>(g)</p><p>CO </p><p>2</p><p> </p><p> </p><p> </p><p> + </p><p>(l)</p><p>2H </p><p>2</p><p> </p><p> O</p><p> </p><p> </p><p>⇒Δn </p><p>g</p><p> </p><p> =1−3=(−2)</p><p>ΔH=ΔU−2RT$

Given, Heat of combustion= 885389kJ/mol

$⇒ΔH=885389− </p><p>1000</p><p>2(8.314)(298)</p><p>$

=(885389−4.95)kJ/mol

∴ Enthalpy is= 885384.05kJ/mol

Answer 2

Answer:

The enthalpy of combustion of CH₄ is 0.105 Joule.

Explanation:

The reaction will proceed as follows,

$CH_4+2O_2 \rightarrow CO_2+2H_2O$     (1)

The enthalpy of combustion at constant pressure is given as,

$\Delta H=nC\Delta T$          (2)

Where,

ΔH=enthalpy at constant pressure

n=number of moles of the substance

C=specific heat

ΔT=change in temperature

From the question we have,

Mass of CH₄=1g

ΔT=$\frac{2}{5}$°C

C=1 cal/gram

The number of moles of CH₄ is given as,

$n=\frac{1}{16}$         (3)       (molar mass of CH₄ is 16)

By substituting all the required values in equation (2) we get;

$\Delta H=\frac{1}{16} \times 1\times \frac{2}{5}$

$\Delta H=0.025 cal$

$\Delta H=0.105 Joule$

Hence, the enthalpy of combustion of CH₄ is 0.105 Joule.