Question

gaseous hydrogen and gaseous chlorine can react to form hydrogen chloride gas, as represented by the following equation

equimolar amounts of ( h_2 ) and ( cl_2 ) are placed in an evacuated reaction vessel at 600 k. which of the following statements correctly predicts how the pressure will be affected as the reaction proceeds at a constant temperature and explains why?

a the pressure will remain constant because the number of gas particles is constant.

b the pressure will be halved because one substance is produced from two substances.

c the pressure will increase because the coefficient of hcl in the balanced chemical equation is greater than that of the reactants.

d the pressure will decrease because the average speed of the gaseous product molecules is less than the average speed of the

Explanation:

Step1: Analyze the gas - particle number

According to the ideal gas law \(PV = nRT\) (at constant \(V\), \(T\), \(R\)), \(P\propto n\) (where \(n\) is the number of moles of gas).
In the reaction \(H_{2}(g)+Cl_{2}(g)\to2HCl(g)\), the number of moles of gas on the reactant side: \(n_{reactants}=n(H_{2}) + n(Cl_{2})\). Since \(n(H_{2})=n(Cl_{2})\) (equimolar amounts), let \(n(H_{2}) = n(Cl_{2})=x\), then \(n_{reactants}=x + x=2x\).
The number of moles of gas on the product side: \(n_{products}=2n(HCl)\). From the stoichiometry of the reaction, if \(x\) moles of \(H_{2}\) and \(x\) moles of \(Cl_{2}\) react, \(n(HCl) = 2x\) (because of the \(1:1:2\) mole ratio in \(H_{2}+Cl_{2}\to2HCl\)), so \(n_{products}=2x\).

Step2: Relate mole number to pressure

Since \(n_{reactants}=n_{products}\) (both are \(2x\) when considering the stoichiometry and equimolar starting amounts of \(H_{2}\) and \(Cl_{2}\)), and \(P\propto n\) (at constant \(V\), \(T\), \(R\)), the pressure will remain constant.

Answer:

A. The pressure will remain constant because the number of gas particles is constant.