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For the synthesis of ammonia$$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$the equilibrium constant $K_{\mathrm{c}}$ at $375^{\circ} \mathrm{C}$ is $1.2 .$ Starting with $\left[\mathrm{H}_{2}\right]_{0}=0.76 \mathrm{M},\left[\mathrm{N}_{2}\right]_{0}=0.60 \mathrm{M},$ and $\left[\mathrm{NH}_{3}\right]_{0}=$$0.48 M,$ which gases will have increased in concentration and which will have decreased in concentration when the mixture comes to equilibrium?

$\left[\mathrm{N}_{2}\right]$ and $\left[\mathrm{H}_{2}\right]$ will decrease $\left[\mathrm{NH}_{3}\right]$ will increase

Chemistry 102

Chapter 14

Chemical Equilibrium

University of Central Florida

Brown University

University of Toronto

Lectures

10:03

In thermodynamics, a state of thermodynamic equilibrium is a state in which a system is in thermal equilibrium with its surroundings. A system in thermodynamic equilibrium is in thermal equilibrium, mechanical equilibrium, electrical equilibrium, and chemical equilibrium. A system is in equilibrium when it is in thermal equilibrium with its surroundings.

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In chemistry, chemical equilibrium (also known as dynamic equilibrium) is a state of chemical stability in which the concentrations of the chemical substances do not change in the course of time due to their reaction with each other in a closed system. Chemical equilibrium is an example of dynamic equilibrium, a thermodynamic concept.

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So for this problem, we want to know the relative concentrations of the products in reactions at equal a room, given our initial values that we received in the question itself. So what we want to do first if we want to calculate q c que si is the reaction quotient, and it tells everything you know about the relation between products and reactions in terms of concentrations at a 0.4 or after equilibrium. So just like what we dio in with calculating the equilibrium constant Casey, we're going to dio concentration off our products. Raised to the second I said So many walls we have here and then we wouldn't you dio nice surgeon, our first reacted multiplied by hydrogen are second reacted. Race had three. That's how many we have here. And we were given these values in the question. So you know, the concentration of our product is zero point for eight squared, and we know that for nitrogen, 0.6 for hydrogen, it is 0.76 So when we calculate this out, we see that the reaction quotient is zero point 87 So we know now that K C is larger thank you. See? So because of this, we know that it will shift right at equilibrium, which means our concentration of in each three at equal a real will increase while our concentrations of N two she's nitrogen and H two hydrogen will decrease.

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