Chapter 6, Problem 6.107P

Interpretation Introduction

Interpretation:

The heat released when $15.0\text{ L}$ of $\text{CO}$ reacts with $18.5\text{ L}$ of ${\text{H}}_2$ is to be determined.

Concept introduction:

The standard enthalpy of reaction is calculated by the summation of standard enthalpy of formation of the product minus the summation of standard enthalpy of formation of reactant at the standard conditions. The formula to calculate the standard enthalpy of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is as follows:

$\Delta H_{\text{rxn}}^{°}=\sum m\Delta H_{\text{f (products)}}^{°}-\sum n\Delta H_{\text{f (reactants)}}^{°}$

Here, m and n are the stoichiometric coefficients of reactants and product in the balanced chemical equation.

The ideal gas equation can be expressed as follows,

$PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

The conversion factor to convert $\text{atm}$ into $\text{kPa}$ is as follows:

$1\text{atm}=101.325\text{kPa}$