Steelers QB Ben Roethlisberger named AFC Offensive Player of the Week

For the second time in just four weeks, a member of the Pittsburgh Steelers offense has been named the AFC Offensive Player of the Week. DeAngelo Williams was given the award after his phenomenal performance against the Washington Redskins in Week 1, and it was Ben Roethlisberger getting the award after his impeccable show of precision against the Kansas City Chiefs in Week 4.

22 completions ✅
300 passing yards ✅
5 touchdowns ✅
152.5 passer rating ✅
AFC Offensive Player of the Week ✅

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— Pittsburgh Steelers (@steelers) October 5, 2016

Looking at the stats, it is pretty incredible what Roethlisberger was able to do against a very stout Chiefs defense. His 5 touchdowns equated to the same amount of incompletions he threw the entire game. His passer rating of 152.5 was near perfect, and he seemed in complete control over the offense from the first quarter until he was replaced by Landry Jones in the waning minutes of the fourth.

Roethlisberger has had a good start to the 2016 season, completing 63.4-percent of his passes, throwing for 1,116 yards, 11 touchdowns and just 4 interceptions. Roethlisberger will have to keep up this high level of play if the Steelers don't just want to beat the New York Jets in Week 5, but to help propel them to a 7th Lombardi trophy in Super Bowl 51.