A student ran the following reaction in the laboratory at 329 K:CH4(g) + CCl4(g) 2CH₂Cl₂(g)When he introduced CH4(g) and CCl4(g) into a 1.00 L evacuated container, so that the initialpartial pressure of CH4 was 0.602 atm and the initial partial pressure of CCl4 was 0.262 atm,he found that the equilibrium partial pressure of CCl4 was 0.219 atm.Calculate the equilibrium constant, Kp, she obtained for this reaction.Кр=?

Answered: A student ran the following reaction in…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

A student ran the following reaction in the laboratory at 329 K:
CH4(g) + CCl4(g) 2CH₂Cl₂(g)
When he introduced CH4(g) and CCl4(g) into a 1.00 L evacuated container, so that the initial
partial pressure of CH4 was 0.602 atm and the initial partial pressure of CCl4 was 0.262 atm,
he found that the equilibrium partial pressure of CCl4 was 0.219 atm.
Calculate the equilibrium constant, Kp, she obtained for this reaction.
Кр
=
?

Expert Solution

Step 1: Given

The balanced chemical equation given in the question is -

CH_4(g) + CCl_4(g) <-> 2CH₂Cl_2(g)

The ICE table is -

	CH_4(g)	+ CCl_4(g)	<->	2CH₂Cl_2(g)
I	0.602 atm	0.262 atm		0 atm
C	(0.602-x) atm	(0.262-x) atm		2x atm
E	(0.602 - x)	(0.262 - x) = 0.219 atm x = 0.262 - 0.219 x = 0.043		2x atm
E	(0.602 - x) = 0.602 - 0.043 =0.559	x=0.043		2x atm = 2 x 0.043 = 0.086

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