Question

date:
name(s):

1. for the reaction below, initially 3.5 mol of nh, are placed in a 4.0 l

reaction chamber. after 3.0 minutes only 1.6 moles of nh, remain.
show all your work:
4nh₃(g)+5o₂(g)→4no(g)+6h₂o(g)
a. calculate the average rate of reaction with respect to nh₃.
b. calculate the average rate at which h₂o is being formed.
c. calculate the average rate at which o₂ is being consumed.
(answer = units for
nh₃)
(answer = units for
h₂o)
(answer = units for o₂)
date:
name(s):
page 5 / 6
2no(g)+br₂(g)→2nobr(g)

Explanation:

Step1: Calculate the change in concentration of \(NH_3\)

The initial moles of \(NH_3,n_{initial}=3.5\ mol\), the final moles \(n_{final} = 1.6\ mol\), and the volume \(V = 4.0\ L\).
The change in concentration \(\Delta[NH_3]=\frac{n_{final}-n_{initial}}{V}=\frac{1.6 - 3.5}{4.0}\ mol/L=- 0.475\ mol/L\)
The time interval \(\Delta t=3.0\ min\)

Step2: Calculate the average rate of reaction with respect to \(NH_3\)

The formula for the average rate of reaction with respect to a reactant \(A\) is \(Rate=-\frac{\Delta[A]}{\Delta t}\)
For \(NH_3\), \(Rate_{NH_3}=-\frac{\Delta[NH_3]}{\Delta t}\)
Substitute \(\Delta[NH_3]=- 0.475\ mol/L\) and \(\Delta t = 3.0\ min\)
\(Rate_{NH_3}=-\frac{-0.475}{3.0}\ mol/(L\cdot min)\approx0.158\ mol/(L\cdot min)\)

Step3: Use stoichiometry to find the rate of formation of \(H_2O\)

From the balanced equation \(4NH_3(g)+5O_2(g)\to4NO(g)+6H_2O(g)\)
The ratio \(\frac{Rate_{H_2O}}{Rate_{NH_3}}=\frac{6}{4}\)
\(Rate_{H_2O}=\frac{6}{4}\times Rate_{NH_3}\)
Substitute \(Rate_{NH_3}=0.158\ mol/(L\cdot min)\)
\(Rate_{H_2O}=\frac{6}{4}\times0.158\ mol/(L\cdot min)=0.237\ mol/(L\cdot min)\)

Step4: Use stoichiometry to find the rate of consumption of \(O_2\)

From the balanced equation, the ratio \(\frac{Rate_{O_2}}{Rate_{NH_3}}=\frac{5}{4}\)
\(Rate_{O_2}=\frac{5}{4}\times Rate_{NH_3}\)
Substitute \(Rate_{NH_3}=0.158\ mol/(L\cdot min)\)
\(Rate_{O_2}=\frac{5}{4}\times0.158\ mol/(L\cdot min)=0.198\ mol/(L\cdot min)\)

Answer:

a. \(0.16\ mol/(L\cdot min)\)
b. \(0.24\ mol/(L\cdot min)\)
c. \(0.20\ mol/(L\cdot min)\)