Question

1. given that when, in the reaction above, 2.000 g of ch4(g) is used in a calorimeter with heat capacity of 2139 j/°c, the temperature falls from 25.26 to 22.88 °c. 6. from tables of δhf°.

Explanation:

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=22.88 - 25.26=- 2.38^{\circ}C$

Step2: Calculate heat released by the reaction

$q = C\times\Delta T$, where $C = 2139\ J/^{\circ}C$. So $q=2139\times(-2.38)= - 5090.82\ J$

Step3: Calculate moles of $CH_4$

The molar mass of $CH_4$ is $M=(12.01 + 4\times1.01)\ g/mol=16.05\ g/mol$. Moles of $CH_4$, $n=\frac{m}{M}=\frac{2.000\ g}{16.05\ g/mol}=0.1246\ mol$

Step4: Calculate $\Delta H$ per mole

$\Delta H=\frac{q}{n}=\frac{-5090.82\ J}{0.1246\ mol}\approx - 40857.3\ J/mol=-40.86\ kJ/mol$

Answer:

The value of $\Delta H$ for the reaction (per mole of $CH_4$) is approximately $- 40.86\ kJ/mol$