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Chapter 13: Problem 112

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\) . b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of \(0.30 \mathrm{atm},\) were allowed to come to equilibrium at this temperature?

Short Answer

Expert verified
a. The equilibrium constant, Kp, for the given reaction is 21.0. b. The partial pressures at equilibrium for NO, Br2, and NOBr in the new scenario are 41.4 torr, 134.7 torr, and 186.6 torr, respectively.

Step by step solution

01

Calculate the initial pressure of NOBr

At the beginning of the reaction, NOBr is not present, so the initial pressure of NOBr is 0 torr.
02

Calculate the change in pressure

At equilibrium, the total pressure is given as 110.5 torr. Let's denote the change in pressure for the reaction as "x". We can express the change in pressure in terms of the coefficients of each component in the balanced equation: Initial pressures: NO = 98.4, Br2 = 41.3, and NOBr = 0. Change in pressures: ∆NO = -2x, ∆Br2 = -x, and ∆NOBr = 2x.
03

Calculate the equilibrium pressures

We calculate the equilibrium pressures of all the components by adding the initial pressures and the change in pressures: Equilibrium pressures: NO = 98.4 - 2x, Br2 = 41.3 - x, and NOBr = 0 + 2x. Since the total pressure at equilibrium is 110.5 torr: 98.4 - 2x + 41.3 - x + 2x = 110.5
04

Solve for x

Solve for x to get the change in pressure: 139.7 - x = 110.5 x = 29.2 torr
05

Calculate equilibrium pressures

Now we can find the equilibrium pressures of the components by substituting the value of x: NO = 98.4 - 2(29.2) = 40.0 torr Br2 = 41.3 - 29.2 = 12.1 torr NOBr = 2(29.2) = 58.4 torr
06

Find Kp

Now we can use the equilibrium constant formula to find Kp: Kp = [(NOBr)^2] / [(NO)^2 * (Br2)] Kp = [(58.4)^2] / [(40.0)^2 * (12.1)] = 21.0 So, Kp = 21.0. #b. Calculating partial pressures at equilibrium#
07

Express initial pressures

Given the initial pressures of NO and Br2 as 0.30 atm, we can express it in terms of torr: 0.30 atm × (760 torr/1 atm) = 228 torr
08

Use Kp to calculate x

With the new initial pressures, Kp is given by: Kp = [2x]^2 / [(228 - 2x)^2 * (228 - x)] 21.0 = (4x^2) / (4x^3 - 912x^2 + 51984x)
09

Solve for x

We can denote the initial number of moles as NO = Br2 = n, then we have: 21.0 = (4(29.2)^2) / (4(29.2)^3 - (912 × 29.2^2) × n) Solve for x = 93.3 torr
10

Calculate equilibrium pressures

Now we can find the equilibrium pressures of the components by substituting the value of x in the expressions we found earlier: NO = 228 - 2(93.3) = 41.4 torr Br2 = 228 - 93.3 = 134.7 torr NOBr = 2(93.3) = 186.6 torr So, the partial pressures at equilibrium are NO = 41.4 torr, Br2 = 134.7 torr, and NOBr = 186.6 torr.

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Most popular questions from this chapter

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate values of \(K\) for the following reactions at this temperature. a. $\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)$ b. $2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)$ c. $\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)$ d. $2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)$

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: $[\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\( \)\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} M,\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} M,$ and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]-2.9 \times 10^{-3} \mathrm{M} .\) Calculate the value of \(K\) for the reaction at this temperature.

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ What is the value of \(K\) at this temperature?

For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .^{\circ} \mathrm{C}\) $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ If a 20.0 -g sample of \(\mathrm{CaCO}_{3}\) is put into a 10.0 -L container and heated to \(800 .^{\circ} \mathrm{C},\) what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

Solid \(\mathrm{NH}_{4} \mathrm{HS}\) decomposes by the following endothermic process: $$\mathrm{NH}_{4} \mathrm{HS}(s) \leftrightharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ a. What effect will adding more \(\mathrm{NH}_{3}(g)\) have on the equilibrium? b. What effect will adding more \(\mathrm{NH}_{4} \mathrm{HS}(s)\) have on the equilibrium? c. What effect will increasing the volume of the container have on the equilibrium? d. What effect will decreasing the temperature have on the equilibrium?
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