The rale constant for the gaseous reaction
H2(g) + I2(g) → 2HI(g)
is 2.42 × 10−2/M · s at 400°C. Initially an equimolar sample of H2 and I2 is placed in a vessel at 400°C, and the total pressure is 1658 mmHg. (a) What is the initial rate (M/min) of formation of HI? (b) What are the rate of formation of HI and the concentration of HI (in molarity) after 10.0 min?
Interpretation:
The initial rate of formation of
Concept introduction:
Rate law: It is an equation that related to the rate of reaction to the concentrations or pressures of substrates (reactants). It is also said to be as rate equation.
Rate: The rate is nothing but the change in concentration of substrate (reactant) or target (product) with time.
- The change in concentration term is divided by the respective stoichiometric coefficient.
- The negative sign indicates that substrates (reactants) concentration decrease as per the reaction progress.
- Rate of reaction is always represented by positive quantities.
Answer to Problem 19.131QP
The initial rate of formation of
The rate formation of
Explanation of Solution
(a)
From the given information, the units of rate constant denotes a second-order reaction and the rate law is probably as follows
By using the ideal gas equation to solve for the initial concentrations of
The total pressure is 1658 mmHg and there are equimolar amounts of
Now, let’s convert the units of the rate constant to /M.min, and substitute the corresponding values in the rate law to solve the rate as follows.
We know that,
or
(b)
The integrated second-order rate law is used to calculate the concentration of
Similarly we can calculate the concentration of
Now,
We know that,
Therefore, the concentration of
The initial rate of formation of
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Chapter 19 Solutions
GEN COMBO CHEMISTRY: ATOMS FIRST; ALEKS 360 2S ACCESS CARD CHEMISTRY:ATOMS FIRST
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