QUESTION IMAGE
Question
- 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)
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 \(NH_3\) (part a)
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: Calculate the average rate at which \(H_2O\) is being formed (part b)
From the balanced chemical equation \(4NH_3(g)+5O_2(g)\to4NO(g)+6H_2O(g)\)
The ratio of rates \(Rate_{H_2O}=\frac{6}{4}Rate_{NH_3}\)
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: Calculate the average rate at which \(O_2\) is being consumed (part c)
From the balanced chemical equation, the ratio of rates \(Rate_{O_2}=\frac{5}{4}Rate_{NH_3}\)
Substitute \(Rate_{NH_3}=0.158\ mol/(L\cdot min)\)
\(Rate_{O_2}=\frac{5}{4}\times0.158\ mol/(L\cdot min)=0.198\ mol/(L\cdot min)\)
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