Chapter 10, Problem 31A

Interpretation Introduction

Interpretation:

When 1.00 g of methane is burned in excess of oxygen, the enthalpy change for the reaction should be predicted.

Concept Introduction:

The amount of heat released or absorbed during the reaction at constant pressure is known as enthalpy change.

Answer to Problem 31A

ΔH for the reaction is -55.51 kJ

Explanation of Solution

The given reaction is shown below:

  ${\text{CH}}_{\text{4}}\left(\text{g}\right){\text{ + 2O}}_{\text{2}}\left(\text{g}\right)\to {\text{CO}}_{\text{2}}\left(\text{g}\right){\text{ + H}}_{\text{2}}\text{O}\left(\text{l}\right)$

The calculation of moles is shown below:

  $\begin{array}{c}\text{moles}\text{of}\text{methane}\text{=}\frac{\text{mass}\text{of}\text{methane}}{\text{molar}\text{mass}\text{of}\text{methane}}\\ \text{=}\frac{\text{1}\text{g}}{\text{16}\text{.04}{\text{gmol}}^{\text{-1}}}\\ \text{=}\text{0}\text{.0623}\text{mol}\end{array}$

According to the reaction:

1 mol of methane = -891 kJmol-1 of enthalpy change

0.0623 mol of methane = -891 kJmol-1× 0.0623 mol of enthalpy change

  = - 55.51 kJ of enthalpy change

Conclusion

Thus, the value of ΔH for the reaction is -55.51 kJ.