The first-order rate constant for the decomposition of N2O5, 2 N2O51g2¡4 NO21g2 + O21g2, a t 70 C i s 6.82 * 10-3 s-1. Suppose we start with 0.0250 mol of N2O51g2 in a volume of 2.0 L. (a) How many moles of N2O5 will remain after 5.0 min?

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All right. Hi, everyone. So this question says that the first order rate constant for the decomposition of the hypothetical compound A two B five in which two moles of A two B five as a gas dissociate to form four moles of A B two and one mole of B two at 70 °C is 2.34 multiplied by 10 to the negative third power inverse seconds. Suppose we start with 0.01250 moles of A two B five in a volume of 1.5 L. How many moles of A two B five will remain after 3.0 minutes? And here we have four different answer choices proposing different values in moles. So let's go ahead and get started. Now, in this case, we're given a specific chemical reaction in which two moles of A two B five as a gas decompose to form four moles of A B two, which is also a gas and one mole of B two, which is a guess. Now the temperature is 70 °C and K which is the rate constant is equal to 2.34 multiplied by 10 to the negative third in verse seconds. Now, in this case, right, we're given the initial concentration or a subzero which is equal to 0.01250 moles of A two B five divided by 1.5 L of solution. So using the first order or rather the integrated Great law for a first order reaction, we can calculate the concentration of A two B five after three minutes have passed. Now because a rate law or rate constant excuse me is given in units of inverse seconds. Recall that 3.0 minutes is equal to 180 seconds. So in this case, the natural logarithm of our compound A after 180 minutes is equal to negative KT added to the natural logarithm of a subzero or the initial concentration. So here we can use the information given to solve for the concentration after 180 seconds. So that's negative 2.34 multiplied by 10 to the negative third power in verse seconds multiplied by T which is 180 seconds. And this is added to the natural logarithm of the initial concentration which is 0.01250 moles divided by 1.5 L. So after evaluating the right side of the equation, the Ln of a sub 180 is equal to negative 5.20 8692. Now to solve for the concentration of A two B five, we're going to make both sides of the equation an exponent of E. So a sub 180 is equal to E to the power of negative 5.208692. And this equals oops, this equals five point four 68822, multiplied by 10 to the negative third moles per liter. However, because the question is asking for the amount in moles, let's take the concentration. It's 5.468822 multiplied by 10 to the negative third moles per liter and multiply it by the volume of the solution which is 1.5 L. This cancels out units of leaders and equals 8.2 multiplied by 10 to the negative third moles after rounding to two significant figures and there you have it. So scrolling up here after 3.0 minutes, there were 8.2 multiplied by 10 to the negative third moles of A two B five which corresponds to option C in the multiple choice. So with that being said, thank you so very much for watching. And I hope you found this helpful.

The first-order rate constant for the decomposition of N2O5, 2 N2O51g2¡4 NO21g2 + O21g2, a t 70 C i s 6.82 * 10-3 s-1. Suppose we start with 0.0250 mol of N2O51g2 in a volume of 2.0 L. (a) How many moles of N2O5 will remain after 5.0 min?

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Key Concepts

First-Order Kinetics

Integrated Rate Law

Molarity and Moles