Question

Problem PageQuestion A chemist fills a reaction vessel with nitrogen monoxide gas, chlorine gas, and nitrosyl chloride gas at a temperature of . Under these conditions, calculate the reaction free energy for the following chemical reaction: Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer 1

The question is incomplete, here is the complete question:

A chemist fills a reaction vessel with 9.47 atm nitrogen monoxide gas, 2.61 atm chlorine gas, and 8.64 atm nitrosyl chloride gas at a temperature of 25°C. Under these conditions, calculate the reaction free energy for the following chemical reaction:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

Answer: The Gibbs free energy change of the reaction is -44.0 kJ

Explanation:

The equation used to calculate standard Gibbs free change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

• The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NOCl(g))})]-[(2\times \Delta G^o_f_{(NO(g))})+(2\times \Delta G^o_f_{(Cl_2)})]$

We are given:

$\Delta G^o_f_{(NOCl(g))}=66.08kJ/mol\\\Delta G^o_f_{(Cl_2(g))}=0kJ/mol\\\Delta G^o_f_{(NO(g))}=86.55kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (66.08))]-[(2\times (86.55))+(1\times (0))]\\\\\Delta G^o_{rxn}=-40.94kJ$

• The expression of $Q_p$ for above equation follows:

$Q_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=8.64atm\\p_{NO}=9.47atm\\p_{Cl_2}=2.61atm$

Putting values in above expression, we get:

$Q_p=\frac{(8.64)^2}{(9.47)^2\times 2.61}=0.319$

To calculate the Gibbs free energy change of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln Q_p$

where,

$\Delta G^o$ = standard Gibbs free energy change = -40.94 kJ

R = Gas constant = 8.314 J/mol.K = 0.008314 kJ/mol.K

T = temperature = $25^oC=[25+273]K=298K$

$Q_p$ = reaction coefficient = 0.319

Putting values in above equation, we get:

$\Delta G=-40.94+(0.008314J/mol.K\times 298K\times \ln (0.319)\\\\\Delta G=-43.77kJ$

Hence, the Gibbs free energy change of the reaction is -44.0 kJ