Question

CHM 202 HW 8 F 19 1. A reaction mixture at 350 C was found to have the following concentrations: [HI] = 0.024 M; [H2] = 0.025 M and [2] = 0.025 M H2(g) + talg) 2 Hig), K. = 57.0 a) Determine whether the mixture is at equilibrium or not b) if the mixture is not at equilibrium, tell the direction of the shift to reach equilibrium, and calculate the equilibrium concentrations of H2, I2 and HL 2. Consider the Haber Process for the synthesis of ammonia from hydrogen and nitrogen 3 H2(g) + Nag) 2NH:(g) Show that the reaction would shift to the right if the pressure is doubled. 3. a) How would you drive the following reaction to the right to form more products: CO(g) + H20(e) = cox)+H; (g) AH°= -40.3 kJ Ans: b) One of the products is a high value substance. Which one, and what is it used for?

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Answer 1

Hi,

Hope you are doing well.

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PART (a):

Given reaction:

$H_{2}_{(g)}+I_{2}_{(g)}\rightarrow 2HI_{(g)}\;\;\;;\mathbf{K_{c}=57}$

Reaction quotient of the given reaction is given by,

$Q=\frac{\left [ HI \right ]^{2}}{\left [ H_{2} \right ].\left [ I_{2} \right ]}=\frac{(0.024\;M)^{2}}{(0.025\;M)\times (0.025\;M)}$

$\mathbf{\therefore Q=0.9216}$

$\mathbf{\therefore Q<K_{c}}$

Since, $\mathbf{ Q\neq K_{c}}$ , the mixture is NOT in equilibrium.

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PART (b):

Since, $\mathbf{Q<K_{c}}$ , forward reaction will happen to reach equilibrium.

From ICE table we have:

At equilibrium we have,

$K_{c}=\frac{\left [ HI \right ]^{2}}{\left [ H_{2} \right ].\left [ I_{2} \right ]}=\frac{(0.024&plus;x)^{2}}{(0.025-x)\times (0.025-x)}=57$

Solving for x, we get

$\therefore (0.024&plus;x)^{2}=57(0.025-x)^{2}$

$\therefore 0.024&plus;x=\sqrt{57}(0.025-x)$

$\therefore x=\frac{\sqrt{57}\times 0.025-0.024}{1&plus;\sqrt{57}}$

$\therefore x=0.01927\;M$

Hence concentration of each molecules at equilibrium is given by,

$\left [ H_{2} \right ]=0.025\;M-x=0.025\;M-0.01927\;M$

$\mathbf{\therefore \left [ H_{2} \right ]=0.00573\;M}$

$\left [ I_{2} \right ]=0.025\;M-x=0.025\;M-0.01927\;M$

$\mathbf{\therefore \left [ I_{2} \right ]=0.00573\;M}$

$\left [ HI \right ]=0.024\;M&plus;x=0.024\;M&plus;0.01927\;M$

$\mathbf{\therefore \left [ HI \right ]=0.0433\;M}$

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