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Chapter 17: Problem 110

The \(K_{P}\) for the reaction $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} $$ is \(2.4 \times 10^{-3}\) at \(720^{\circ} \mathrm{C}\). What is the minimum partial pressure of \(\mathrm{N}_{2}\) required for the reaction to be spontaneous in the forward direction if the partial pressures of \(\mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are \(1.52 \mathrm{~atm}\) and \(2.1 \times 10^{-2} \mathrm{~atm}\) respectively?

Short Answer

Expert verified
By calculating the equation, the minimum partial pressure of nitrogen required for the reaction to be spontaneous in the forward direction is greater than 0.281 atm.

Step by step solution

01

Calculate the Reaction Quotient (Q)

First, the reaction quotient \(Q\) needs to be calculated. This can be done using the formula \(Q_{p} = \frac{([NH_3]^2)}{([N_2][H_2]^3)}\), where the pressures are replaced with their respective values. Here, the pressure of \(NH_{3}\) is given as \(2.1 \times 10^{-2} atm\) and the pressure of \(H_{2}\) as \(1.52 atm\). However, the pressure of \(N_{2}\) is what needs to be found, so for simplicity's sake denote it as \(x\). Thus, the equation becomes \(Q_{p} = \frac{[(2.1 \times 10^{-2})^2]}{[x(1.52)^3]}\)
02

Set Q less than Kp

We know that for the reaction to be spontaneous in the forward direction, the reaction quotient Q must be less than the equilibrium constant \(K_{p}\). The equilibrium constant \(K_{p}\) is given as \(2.4 \times 10^{-3}\). So, now set up an inequality where \(Q_{p} < K_{p}\) becomes \(\frac{[(2.1 \times 10^{-2})^2]}{[x(1.52)^3]} < 2.4 \times 10^{-3}\)
03

Solve for N2 pressure (x)

Now, we have to solve this equation for \(x\) which represents the Nitrogen's partial pressure. By doing the calculations and isolating \(x\) on one side of the equation, we get \(x > \frac{(2.1 \times 10^{-2})^2}{(2.4 \times 10^{-3})(1.52)^3}\). This will give us the partial pressure of the Nitrogen gas.

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Most popular questions from this chapter

A student placed \(1 \mathrm{~g}\) of each of three compounds A, B, and C in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s),\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g) .\) Give the reasons for your arrangement.

For reactions carried out under standard-state conditions, Equation (17.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}\) \(-T \Delta S^{\circ}\). (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$ \ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right) $$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C}, K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\).

For a reaction with a negative \(\Delta G^{\circ}\) value, which of the following statements is false? (a) The equilibrium constant \(K\) is greater than one. (b) The reaction is spontaneous when all the reactants and products are in their standard states. (c) The reaction is always exothermic.

Consider the decomposition of calcium carbonate: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ Calculate the pressure in atm of \(\mathrm{CO}_{2}\) in an equilibrium process (a) at \(25^{\circ} \mathrm{C}\) and \((\mathrm{b})\) at \(800^{\circ} \mathrm{C}\). Assume that \(\Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S^{\circ}=160.5 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) for the temperature range.
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