Answer
The molar concentration of $NOBr$ is $1.4 \space M$
Work Step by Step
1. Write the equilibrium constant expression:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ NO ]^{ 2 }[ Br_2 ]}{[ NOBr ]^{ 2 }}$$
2. Solve for the missing concentration:
$$ [NOBr] = \sqrt[2]{\frac{[ NO ]^{ 2 }[ Br_2 ]}{ K_c}}$$
3. Evaluate the expression:
$$ [NOBr] = \sqrt[2]{\frac{( 2.0 )^{ 2 }( 1.0 )}{(2.0)}}$$
$$[NOBr] = 1.4 \space M$$
